A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If ${\rm{\mu }}$ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is

${\cos ^{ - 1}}\,\,\left( {\frac{1}{{\rm{\mu }}}} \right)$

${\sin ^{ - 1}}\,\,\left( {\frac{1}{{\rm{\mu }}}} \right)$

${\tan ^{ - 1}}\,\,\left( {\frac{1}{{\rm{\mu }}}} \right)$

${\sin ^{ - 1}}\,\,\left( {\frac{{{\rm{\mu }} - 1}}{{\rm{\mu }}}} \right)$

An astronomical telescope has an angular magnification of magnitude ${\rm{5}}$ for distant objects. The separation between the objective and the eye-piece is ${\rm{36}}\,\,{\rm{cm}}$ and the final image is formed at infinity. The focal length ${f_0}$ of the objective and the focal length ${f_e}$ of the eye-piece are

${f_0} = 30\;\,{\rm{cm\;and\;}}\,{f_e} = 6\,\;{\rm{cm}}$

${f_0} = 45\;\,{\rm{cm\;and\;}}\,{f_e} = - \,9\,\;{\rm{cm}}$

${f_0} = - 7.2\;\,{\rm{cm\;and\;}}\,{f_e} = 5\;\,{\rm{cm}}$

${f_0} = 50\,\;{\rm{cm\;and\;}}\,{f_e} = 10\;\,{\rm{cm}}$

A person cannot see properly beyond ${\rm{2}}\,\,{\rm{m}}$. Power of the lens is

${\rm{ - 0}}{\rm{.5}}\,\,{\rm{D}}$

${\rm{0}}{\rm{.5}}\,\,{\rm{D}}$

${\rm{1}}{\rm{.5}}\,\,{\rm{D}}$

${\rm{ - 2}}{\rm{.5}}\,\,{\rm{D}}$

RAY OPTICS MCQs for NEET UG PHYSICS — Mixed difficulty

1 Medium

1 of 241Question permalink

Application +4 | -1

An object 5 cm tall is placed

\mathrm{1}\,\,\mathrm{m}

from a concave spherical mirror which has a radius of curvature of

\mathrm{20}\,\,\,\mathrm{cm}

. The size of the image is

View answer and explanation

Correct option: C

\mathrm{0}\mathrm{.55}\,\,\mathrm{cm}

Option C (

\mathrm{0}\mathrm{.55}\,\mathrm{cm}

) is correct because the size of the image is calculated using the magnification formula for a concave mirror.

Solution approach

Correct Answer + Core Reason: $\mathrm{0}\mathrm{.55}\,\mathrm{cm}$ . The size of the image is calculated using the mirror formula and magnification.

Step-by-Step Solution:

Step 1: Given: Object height $h_o = 5 \, \mathrm{cm}$ , object distance $u = -1 \, \mathrm{m} = -100 \, \mathrm{cm}$ , radius of curvature $R = -20 \, \mathrm{cm}$ .

Step 2: The focal length $f = \frac{R}{2} = \frac{-20}{2} = -10 \, \mathrm{cm}$ .

Step 3: Using the mirror formula, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ , we have $\frac{1}{-10} = \frac{1}{v} + \frac{1}{-100}$ .

Step 4: Solving for $v$ , we get $\frac{1}{v} = \frac{1}{-10} + \frac{1}{100} = \frac{-10 + 1}{100} = \frac{-9}{100}$ . Thus, $v = -\frac{100}{9} \, \mathrm{cm}$ .

Step 5: The magnification $m = -\frac{v}{u} = -\frac{-\frac{100}{9}}{-100} = -\frac{1}{9}$ .

Step 6: The image height $h_i = m \times h_o = -\frac{1}{9} \times 5 = -\frac{5}{9} \, \mathrm{cm} \approx -0.55 \, \mathrm{cm}$ . The size of the image is $0.55 \, \mathrm{cm}$ .

Why other options are wrong:
- Option A: This is incorrect because it is not the correct calculation of the image size using the mirror formula and magnification.
- Option B: This is incorrect because it is not the correct calculation of the image size using the mirror formula and magnification.
- Option D: This is incorrect because it is not the correct calculation of the image size using the mirror formula and magnification.

Key Concepts:
- Mirror formula
- Magnification
- Concave mirror

Memory Aids:
- Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
- Magnification: $m = -\frac{v}{u} = \frac{h_i}{h_o}$

Quick Tips:
- Use the correct sign conventions for object and image distances and heights.
- Remember that a negative magnification indicates an inverted image.

2 Medium

2 of 241Question permalink

Understanding +4 | -1

Red colour is used for danger signals because

View answer and explanation

Correct option: B

It undergoes least scattering

Option B (It undergoes least scattering) is correct because red light has the longest wavelength and is scattered the least by atmospheric particles, making it visible from a greater distance.

Solution approach

Correct Answer + Core Reason: It undergoes least scattering. Red light has the longest wavelength and is scattered the least.

Step-by-Step Solution:

Step 1: Recall Rayleigh's scattering law: Scattering is inversely proportional to the fourth power of the wavelength ( $\lambda$ ). $Scattering \propto \frac{1}{{{\lambda^4}}}$ .

Step 2: Compare wavelengths of different colors: Red light has the longest wavelength in the visible spectrum.

Step 3: Apply Rayleigh's scattering law: Since red light has the longest wavelength, it experiences the least scattering.

Step 4: Conclude why red is used for danger signals: Because it is scattered the least, red light can be seen from a greater distance, making it suitable for danger signals.

Why other options are wrong:
- Option A: Red color does not inherently cause fear, although it can be associated with danger.
- Option C: Red light undergoes the least scattering, not the maximum.
- Option D: While there might be international conventions, the primary reason is the low scattering of red light.

Key Concepts:
- Rayleigh Scattering
- Wavelength of Light
- Visible Spectrum

Memory Aids:
- Red light has the longest wavelength and scatters the least.

Quick Tips:
- Remember that scattering is inversely proportional to the fourth power of the wavelength.

3 Medium

3 of 241Question permalink

Analyzing +4 | -1

A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If

\mathrm{\mu }

is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is

View answer and explanation

Correct option: D

{\cos ^{ - 1}}\,\,\left( {\frac{1}{\mathrm{\mu }}} \right)

Option D (

{\cos ^{ - 1}}\,\left( {\frac{1}{\mathrm{\mu }}} \right)

) is correct because it represents the maximum base angle for total internal reflection to occur at the hypotenuse.

Solution approach

Correct Answer + Core Reason: ${\cos ^{ - 1}}\,\left( {\frac{1}{\mathrm{\mu }}} \right)$ . This is the maximum base angle for total internal reflection at the hypotenuse.

Step-by-Step Solution:

Step 1: Let the base angle of the right-angled prism be $\theta$ . Since the ray travels parallel to the base inside the prism, the angle of incidence at the hypotenuse is also $\theta$ .

Step 2: For total internal reflection to occur at the hypotenuse, the angle of incidence $\theta$ must be greater than the critical angle $C$ .

Step 3: The critical angle is given by $\sin C = 1/\mu$ , where $\mu$ is the refractive index of the prism.

Step 4: Therefore, $\theta > C$ , which means $\sin \theta > \sin C = 1/\mu$ .\nHowever, we need to find the *maximum* value of $\theta$ for which total internal reflection occurs. The angle of incidence on the hypotenuse is equal to the base angle. The angle of refraction at the hypotenuse must be 90 degrees for the critical angle condition. The angle of incidence is equal to the base angle. So, the base angle must be less than or equal to the critical angle's complement.

Step 5: Let $\alpha$ be the angle of incidence on the hypotenuse. Then $\alpha = 90^{\circ} - \theta$ . For total internal reflection, $\alpha \ge C$ , where $C$ is the critical angle.\nSo, $90^{\circ} - \theta \ge \sin^{-1}(1/\mu)$ .
$\theta \le 90^{\circ} - \sin^{-1}(1/\mu) = \cos^{-1}(1/\mu)$ .

Why other options are wrong:
- Option A: ${\sin ^{ - 1}}\,\left( {\frac{1}{\mathrm{\mu }}} \right)$ is incorrect; it represents the critical angle itself, not the maximum base angle.
- Option B: ${\tan ^{ - 1}}\,\left( {\frac{1}{\mathrm{\mu }}} \right)$ is incorrect; it's a tangent function and does not relate to the base angle directly.
- Option C: ${\sin ^{ - 1}}\,\left( {\frac{{\mathrm{\mu } - 1}}{\mathrm{\mu }}} \right)$ is incorrect; it's not related to the given information or the prism's properties.

Key Concepts:
- Total Internal Reflection
- Critical Angle
- Geometry of a Prism

Memory Aids:
- Remember that for total internal reflection, the angle of incidence must be greater than the critical angle.

Quick Tips:
- Always consider the geometry of the prism and the angles involved.

4 Medium

4 of 241Question permalink

Analyzing +4 | -1

Statement-A: The minimum distance between an object and its real image formed by a convex lens is

2{f}

.

Statement-B: The distance between an object and its real image is minimum when its magnification is -1.

View answer and explanation

Correct option: D

A is wrong and B is correct

Option D (A is wrong and B is correct) is correct because the minimum distance between an object and its real image formed by a convex lens is 4f, and this occurs when the magnification is -1.

Solution approach

Correct Answer + Core Reason: A is wrong and B is correct. The minimum distance is 4f, and this occurs when magnification is -1.

Step-by-Step Solution:

Step 1: Analyze Statement A: The minimum distance between an object and its real image formed by a convex lens is $2f$ . This statement is incorrect. The minimum distance is actually $4f$ .

Step 2: Analyze Statement B: The distance between an object and its real image is minimum when its magnification is -1. This statement is correct. Magnification of -1 implies that the object and image are the same size and inverted, which occurs when the object is at $2f$ and the image is also at $2f$ , resulting in a minimum distance of $4f$ .

Step 3: Combine the analysis: Since statement A is wrong and statement B is correct, the answer is option D.

Why other options are wrong:
- Option A: A and B are correct: This is incorrect because statement A is false.
- Option B: A and B are wrong: This is incorrect because statement B is true.
- Option C: A is correct and B is wrong: This is incorrect because statement A is false and statement B is true.

Key Concepts:
- Convex lenses and image formation
- Magnification
- Object and image distances

Memory Aids:
- Minimum distance between object and real image for a convex lens is 4f.
- Magnification of -1 corresponds to object at 2f.

Quick Tips:
- Remember the relationship between object distance, image distance, and focal length for convex lenses.
- Understand the meaning of magnification.

5 Medium

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Analyzing +4 | -1

Assertion: We cannot produce a real image by plane or convex mirrors under any circumstances.
Reason: The focal length of a convex mirror is always taken negative.

View answer and explanation

Correct option: D

If both Assertion and Reason are false

Option D (If both Assertion and Reason are false) is correct because both statements are incorrect. Real images can be formed by concave mirrors, and the focal length of a convex mirror is positive.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are false. Both the assertion and the reason are incorrect statements.

Step-by-Step Solution:

Step 1: Analyze the Assertion: A real image can be formed by a concave mirror when the object is placed beyond the focal point. Therefore, the assertion is false.

Step 2: Analyze the Reason: The focal length of a convex mirror is always positive, not negative. Therefore, the reason is false.

Step 3: Since both the assertion and the reason are false, the correct option is D.

Why other options are wrong:
- Option A: If both Assertion and Reason are true and Reason is the correct explanation of Assertion: Incorrect because both are false.
- Option B: If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion: Incorrect because both are false.
- Option C: If Assertion is true but Reason is false: Incorrect because both are false.

Key Concepts:
- Reflection by Mirrors
- Concave and Convex Mirrors
- Real and Virtual Images
- Focal Length

Memory Aids:
- Concave mirrors can form real or virtual images, depending on the object's position.
- Convex mirrors always form virtual, erect, and diminished images.
- Focal length of a convex mirror is positive.

Quick Tips:
- Remember the sign conventions for mirrors.
- Understand the image formation by different types of mirrors.

6 Medium

6 of 241Question permalink

Remembering +4 | -1

If both the object and image are at infinite distance from a refracting telescope its magnifying power will be equal to

View answer and explanation

Correct option: C

The ratio of the focal length of the objective and eyepiece

Option C (The ratio of the focal length of the objective and eyepiece) is correct because the magnifying power of a refracting telescope is the ratio of the objective's focal length to the eyepiece's focal length.

Solution approach

Correct Answer + Core Reason: The ratio of the focal length of the objective and eyepiece. Magnifying power is the ratio of objective to eyepiece focal lengths.

Step-by-Step Solution:

Step 1: Recall the formula for the magnifying power of a refracting telescope.\nThe magnifying power (M) of a refracting telescope is given by:\nM = $\frac{f_o}{f_e}$ , where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.

Step 2: Analyze the given condition.\nThe object and image are at infinite distance, which is the normal adjustment for a telescope.

Step 3: Final Answer: The ratio of the focal length of the objective and eyepiece

Why other options are wrong:
- Option A (The sum of the focal lengths of the objective and the eyepiece): This is incorrect because the magnifying power is a ratio, not a sum.
- Option B (The different of the focal lengths of the two lenses): This is incorrect because the magnifying power is a ratio, not a difference.
- Option D (The ratio of the focal length of the eyepiece and objective): This is incorrect because the objective focal length is in the numerator, and the eyepiece focal length is in the denominator.

Key Concepts:
- Refracting telescope
- Magnifying power
- Objective lens
- Eyepiece

Memory Aids:
- M = fo/fe (Magnification is objective over eyepiece).

Quick Tips:
- Remember the correct order of focal lengths in the magnifying power formula.

7 Medium

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Analyzing +4 | -1

Assertion (A): In Snell's law

\mu=\frac{\sin i}{\sin r}

, angle

i

is always angle of incidence.

Reason (R):

\mu=\frac{\sin i}{\sin r}

is applicable one medium in air (or vacuum).

View answer and explanation

Correct option: C

If assertion is true but reason is false.

Option C (If assertion is true but reason is false) is correct because Snell's law applies to any two media, not just air or vacuum.

Solution approach

Correct Answer + Core Reason: If assertion is true but reason is false. Snell's law applies to any two media, not just air or vacuum.

Step-by-Step Solution:

Step 1: Understand Snell's Law: Snell's law states that $\mu = \frac{\sin i}{\sin r}$ , where $i$ is the angle of incidence and $r$ is the angle of refraction.

Step 2: Analyze the Assertion: The assertion states that in Snell's law, angle $i$ is always the angle of incidence, which is true.

Step 3: Analyze the Reason: The reason states that Snell's law is applicable only when one medium is air or vacuum, which is false. Snell's law applies to any two media.

Why other options are wrong:
- Option A: The reason is false.
- Option B: The reason is false.
- Option D: The assertion is true.

Key Concepts:
- Snell's Law
- Angle of Incidence
- Angle of Refraction

Memory Aids:
- Remember that Snell's law applies to any two media.

Quick Tips:
- Understand the general applicability of Snell's law.

8 Medium

8 of 241Question permalink

Application +4 | -1

Two thin lenses, one of focal length

+ \,60\,\,cm

and the other of focal length

- \,20\,\,cm

are put in contact. The combined focal length is

View answer and explanation

Correct option: D

- \,30\,\,cm

Option D (

\ - \,30\,\,cm

) is correct because it is calculated using the formula for the combined focal length of two lenses in contact.

Solution approach

Correct Answer + Core Reason: $- \,30\,\,cm$ . The combined focal length is calculated using the lens combination formula.

Step-by-Step Solution:

Step 1: Identify the given information:\nFocal length of lens 1, $f_1 = +60\,\,cm$ \nFocal length of lens 2, $f_2 = -20\,\,cm$

Step 2: Recall the formula for the combined focal length ( $F$ ) of two thin lenses in contact:
$\frac{{1}}{{F}} = \frac{{1}}{{f_1}} + \frac{{1}}{{f_2}}$

Step 3: Substitute the given values into the formula:
$\frac{{1}}{{F}} = \frac{{1}}{{60}} + \frac{{1}}{{(-20)}}$

Step 4: Simplify the equation:
$\frac{{1}}{{F}} = \frac{{1}}{{60}} - \frac{{3}}{{60}}$
$\frac{{1}}{{F}} = \frac{{-2}}{{60}}$
$\frac{{1}}{{F}} = \frac{{-1}}{{30}}$

Step 5: Solve for $F$ :
$F = -30\,\,cm$

Final Answer: The combined focal length is $- 30\,\,cm$ .

Why other options are wrong:
- Option A: $+ \,15\,\,cm$ : This is incorrect. It might result from incorrectly adding the focal lengths or using a wrong sign convention.
- Option B: $- \,15\,\,cm$ : This is incorrect. It might result from incorrectly adding the focal lengths or using a wrong sign convention.
- Option C: $+ \,30\,\,cm$ : This is incorrect. It might result from incorrectly adding the focal lengths or using a wrong sign convention.

Key Concepts:
- Combination of lenses
- Focal length

Memory Aids:
- Remember the formula: $\frac{{1}}{{F}} = \frac{{1}}{{f_1}} + \frac{{1}}{{f_2}}$

Quick Tips:
- Pay attention to the signs of the focal lengths (positive for converging, negative for diverging).

9 Medium

9 of 241Question permalink

Understanding +4 | -1

A liquid wets a solid. The meniscus of the liquid in a long vertical capillary tube of that solid is

View answer and explanation

Correct option: B

Concave upward

Option B (Concave upward) is correct because a liquid that wets a solid has adhesive forces greater than cohesive forces, causing the liquid to rise and form a concave meniscus.

Solution approach

Correct Answer + Core Reason: Concave upward. Wetting implies adhesive forces are stronger than cohesive forces, leading to a concave meniscus.

Step-by-Step Solution:

Step 1: Define wetting: A liquid is said to wet a solid surface when the adhesive forces between the liquid and solid molecules are stronger than the cohesive forces within the liquid itself.

Step 2: Consider the forces acting on the liquid at the point of contact with the solid in a capillary tube. The adhesive forces pull the liquid upwards along the solid surface, while the cohesive forces try to keep the liquid molecules together.

Step 3: When adhesive forces are dominant, the liquid molecules are more attracted to the solid than to each other. This causes the liquid to spread out along the solid surface and rise up the capillary tube.

Step 4: The shape of the liquid surface (meniscus) is determined by the balance of these forces. Since the liquid is pulled upwards along the edges of the tube due to adhesion, the meniscus curves upwards, forming a concave shape.

Step 5: Therefore, a liquid that wets a solid will have a concave upward meniscus in a capillary tube.

Why other options are wrong:
- Option A: Flat is incorrect because a flat meniscus occurs when adhesive and cohesive forces are balanced, which is not the case for a wetting liquid.
- Option C: Convex upward is incorrect because a convex meniscus occurs when cohesive forces are stronger than adhesive forces (non-wetting liquid).
- Option D: Cylindrical upward is incorrect because the meniscus is curved, not cylindrical.

Key Concepts:
- Adhesion and cohesion
- Capillary action
- Meniscus formation

Memory Aids:
- Wetting = Concave (think of water on glass)
- Non-wetting = Convex (think of mercury on glass)

Quick Tips:
- Remember that wetting is related to adhesion being stronger than cohesion.

10 Medium

10 of 241Question permalink

Remembering +4 | -1

Which of the following colours suffers maximum deviation in a prism

View answer and explanation

Correct option: B

Blue

Option B (Blue) is correct because blue light has a shorter wavelength and experiences greater deviation in a prism.

Solution approach

Correct Answer + Core Reason: Blue. Blue light has a shorter wavelength and thus experiences maximum deviation in a prism.

Step-by-Step Solution:

Step 1: The deviation of light in a prism depends on the refractive index of the prism material and the wavelength of the light.

Step 2: Shorter wavelengths of light experience greater deviation.

Step 3: The order of wavelengths in visible light is: Violet < Blue < Green < Yellow < Orange < Red.

Step 4: Among the given options, blue light has the shortest wavelength, so it will suffer maximum deviation.

Why other options are wrong:
- Option A: Yellow. This is incorrect because yellow light has a longer wavelength than blue light.
- Option C: Green. This is incorrect because green light has a longer wavelength than blue light.
- Option D: Orange. This is incorrect because orange light has a longer wavelength than blue light.

Key Concepts:
- Deviation of light in a prism
- Wavelength of light
- Refractive index

Memory Aids:
- Remember that shorter wavelengths deviate more in a prism.

Quick Tips:
- Understand the relationship between wavelength and deviation.

11 Hard

11 of 241Question permalink

Application +4 | -1

The diameter of objective lens of a telescope is

6 \mathrm{~cm}

and wavelength of light used is

540 \mathrm{~nm}.

The resolving power of telescope is

View answer and explanation

Correct option: A

9.1 \times 10^4 \mathrm{~rad}^{-1}

Option A (

9.1 \times 10^4 \mathrm{~rad}^{-1}

) is correct. Resolving power is calculated using the formula

1.22 \frac{\lambda}{D}

and then taking the reciprocal.

Solution approach

Correct Answer + Core Reason: $9.1 \times 10^4 \mathrm{~rad}^{-1}$ . Resolving power is the reciprocal of the resolution limit.

Step-by-Step Solution:

Step 1: Identify the given information.\nDiameter of objective lens, $D = 6 \text{ cm} = 0.06 \text{ m}$ \nWavelength of light, $\lambda = 540 \text{ nm} = 540 \times 10^{-9} \text{ m}$

Step 2: Calculate the limit of resolution.
$\theta = 1.22 \frac{\lambda}{D} = 1.22 \times \frac{540 \times 10^{-9}}{0.06} = 1.22 \times 9 \times 10^{-6} = 10.98 \times 10^{-6} \text{ rad}$

Step 3: Calculate the resolving power.
$\text{Resolving Power} = \frac{1}{\theta} = \frac{1}{10.98 \times 10^{-6}} = \frac{10^6}{10.98} \approx 0.091 \times 10^6 = 9.1 \times 10^4 \text{ rad}^{-1}$

Final Answer: $9.1 \times 10^4 \text{ rad}^{-1}$

Why other options are wrong:
- **Option B ( $10^5 \mathrm{~rad}^{-1}$ ): Incorrect calculation of resolving power.
- Option C ( $3 \times 10^4 \mathrm{~rad}^{-1}$ ): Incorrect calculation of resolving power.
- Option D (None of these): The correct answer is among the options.

Key Concepts:
- Resolving power of a telescope
- Rayleigh criterion

Memory Aids:
- Resolving power is the inverse of the limit of resolution.

Quick Tips:**
- Ensure consistent units before calculations.
- Remember the factor 1.22 in the Rayleigh criterion.

12 Medium

12 of 241Question permalink

Application +4 | -1

An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm. The final image is formed at infinity. The focal lengths

{f_0}

of the objective and

{f_e}

of the eyepiece are

View answer and explanation

Correct option: D

30 cm and 6 cm respectively

Option D (30 cm and 6 cm respectively) is correct because the angular magnification is the ratio of the focal lengths, and the separation is the sum of the focal lengths when the final image is at infinity.

Solution approach

Correct Answer + Core Reason: 30 cm and 6 cm respectively. The magnification is the ratio of the focal lengths, and the separation is the sum of the focal lengths.

Step-by-Step Solution:

Step 1: Identify the given information: Angular magnification $M = 5$ , separation between lenses $L = 36 \text{{ cm}}$ , final image at infinity.

Step 2: Recall the formulas for an astronomical telescope with the final image at infinity: $M = \frac{f_o}{f_e}$ and $L = f_o + f_e$ , where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.

Step 3: Set up the equations: $5 = \frac{f_o}{f_e}$ and $36 = f_o + f_e$

Step 4: Solve the equations: From the first equation, $f_o = 5f_e$ . Substitute this into the second equation: $36 = 5f_e + f_e = 6f_e$ . Therefore, $f_e = \frac{36}{6} = 6 \text{{ cm}}$ .

Step 5: Calculate $f_o$ : $f_o = 5f_e = 5 \times 6 = 30 \text{{ cm}}$ .

Why other options are wrong:
- Option A: 45 cm and -9 cm respectively is incorrect because the focal length cannot be negative and the sum is not 36.
- Option B: 50 cm and 10 cm respectively is incorrect because the ratio is not 5 and the sum is not 36.
- Option C: 7.2 cm and 5 cm respectively is incorrect because the ratio is not 5 and the sum is not 36.

Key Concepts:
- Astronomical Telescope
- Angular Magnification

Memory Aids:
- For a telescope at infinity: $M = \frac{f_o}{f_e}$ and $L = f_o + f_e$

Quick Tips:
- Remember the formulas for magnification and length of a telescope when the final image is at infinity.

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13 of 241Question permalink

Remembering +4 | -1

The refractive index of a piece of transparent quartz is the greatest for

View answer and explanation

Correct option: B

Violet light

Option B (Violet light) is correct because the refractive index is inversely proportional to the wavelength of light, and violet light has the shortest wavelength.

Solution approach

Correct Answer + Core Reason: Violet light. Refractive index is inversely proportional to wavelength, and violet light has the shortest wavelength.

Step-by-Step Solution:

Step 1: The refractive index $(\mu)$ of a material depends on the wavelength $(\lambda)$ of light. Generally, the refractive index is higher for shorter wavelengths.

Step 2: The order of wavelengths for visible light is: Violet < Blue < Green < Yellow < Orange < Red.

Step 3: Since violet light has the shortest wavelength, it will have the greatest refractive index.

Why other options are wrong:
- Option A: Red light. This is incorrect because red light has the longest wavelength and therefore the smallest refractive index.
- Option C: Green light. This is incorrect because green light has a longer wavelength than violet light.
- Option D: Yellow light. This is incorrect because yellow light has a longer wavelength than violet light.

Key Concepts:
- Refractive index
- Wavelength of light
- Dispersion

Memory Aids:
- Remember that violet light has the shortest wavelength and the highest refractive index.

Quick Tips:
- Understand the relationship between refractive index and wavelength.

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Application +4 | -1

An astronomical telescope has an angular magnification of magnitude

\mathrm{5}

for distant objects. The separation between the objective and the eye-piece is

\mathrm{36}\,\,\mathrm{cm}

and the final image is formed at infinity. The focal length

{f_0}

of the objective and the focal length

{f_e}

of the eye-piece are

View answer and explanation

Correct option: D

{f_0} = 30\;\,\mathrm{cm\;and\;}\,{f_e} = 6\,\;\mathrm{cm}

Option D (

{f_0} = 30\;\,\mathrm{cm\;and\;}\,{f_e} = 6\;\,\mathrm{cm}

) is correct because it satisfies both the angular magnification and the separation conditions for an astronomical telescope.

Solution approach

Correct Answer + Core Reason: ${f_0} = 30\;\,\mathrm{cm\;and\;}\,{f_e} = 6\;\,\mathrm{cm}$ . This solution satisfies both the magnification and length constraints of the telescope.

Step-by-Step Solution:

Step 1: Identify the given information.\nAngular magnification, $M = 5$ \nSeparation between objective and eyepiece, $L = 36\;\mathrm{cm}$ \nFinal image is formed at infinity.

Step 2: Write the relevant formulas.\nFor an astronomical telescope, the angular magnification when the final image is at infinity is given by:
$M = \frac{{f_0}}{{f_e}}$ \nThe separation between the objective and eyepiece is given by:
$L = f_0 + f_e$

Step 3: Solve the equations.\nWe have two equations:
1. $5 = \frac{{f_0}}{{f_e}}$
2. $36 = f_0 + f_e$ \nFrom equation (1), $f_0 = 5f_e$ \nSubstitute this into equation (2):
$36 = 5f_e + f_e$
$36 = 6f_e$
$f_e = 6\;\mathrm{cm}$ \nNow, find $f_0$ :
$f_0 = 5 \times 6 = 30\;\mathrm{cm}$

Step 4: Final Answer
$f_0 = 30\;\mathrm{cm}$ and $f_e = 6\;\mathrm{cm}$

Why other options are wrong:
- Option A: ${f_0} = 45\;\,\mathrm{cm\;and\;}\,{f_e} = - \,9\;\mathrm{cm}$ : This gives a negative focal length for the eyepiece, which is not physically realistic for a standard telescope setup. Also, $f_0 + f_e = 45 - 9 = 36$ , which satisfies the length condition, but the magnification is $45/9 = 5$ . However, the negative sign is incorrect.
- Option B: ${f_0} = - 7.2\;\,\mathrm{cm\;and\;}\,{f_e} = 5\;\mathrm{cm}$ : This gives a negative focal length for the objective, which is not physically realistic. Also, $f_0 + f_e = -7.2 + 5 = -2.2$ , which does not satisfy the length condition.
- Option C: ${f_0} = 50\;\mathrm{cm\;and\;}\,{f_e} = 10\;\mathrm{cm}$ : This gives $f_0 + f_e = 50 + 10 = 60$ , which does not satisfy the length condition of 36 cm. The magnification is $50/10 = 5$ , which is correct, but the length is wrong.

Key Concepts:
- Astronomical telescope
- Angular magnification
- Focal length

Memory Aids:
- Magnification of telescope: $M = \frac{{f_0}}{{f_e}}$
- Length of telescope (image at infinity): $L = f_0 + f_e$

Quick Tips:
- Remember the sign conventions for focal lengths.
- Always check if the calculated values satisfy both the magnification and length conditions.

15 Medium

15 of 241Question permalink

Understanding +4 | -1

When light enters water from the vacuum, then the wavelength of light

View answer and explanation

Correct option: A

Decreases

Option A is correct because when light enters a medium with a higher refractive index (like water from a vacuum), its speed decreases, and since the frequency remains constant, the wavelength must decrease.

Solution approach

Correct Answer + Core Reason: Decreases. The speed of light decreases in water, and frequency remains constant.

Step-by-Step Solution:

Step 1: Recall the relationship between speed ( $v$ ), frequency ( $f$ ), and wavelength ( $\lambda$ ): $v = f\lambda$ .

Step 2: When light enters water from a vacuum, its speed decreases because water has a higher refractive index than a vacuum.

Step 3: The frequency of light remains constant as it transitions between media.

Step 4: Since $v$ decreases and $f$ remains constant, $\lambda$ must also decrease to maintain the relationship $v = f\lambda$ .

Why other options are wrong:
- Option B: The wavelength increases only when the speed of light increases, which doesn't happen when light enters water from a vacuum.
- Option C: The wavelength does not remain constant because the speed of light changes.
- Option D: The wavelength does not become zero. It decreases, but it still has a finite value.

Key Concepts:
- Speed of light
- Frequency
- Wavelength
- Refractive index

Memory Aids:
- Remember that frequency is a property of the light source and doesn't change when the medium changes.

Quick Tips:
- When light enters a denser medium, its speed and wavelength decrease, but its frequency remains constant.

16 Medium

16 of 241Question permalink

Analyzing +4 | -1

A short straight object of height 100 cm lies before the central axis of a spherical mirror, whose focal length has absolute value

f = 40\,cm

. The image of object produced by the mirror is of height 25 cm and has the same orientation of the object. One may conclude from the information.

View answer and explanation

Correct option: D

Image is virtual, opposite side of convex mirror

Option D (Image is virtual, opposite side of convex mirror) is correct because a convex mirror always produces virtual, erect, and diminished images.

Solution approach

Correct Answer + Core Reason: Image is virtual, opposite side of convex mirror. A convex mirror always forms a virtual, erect, and diminished image.

Step-by-Step Solution:

Step 1: The object is straight and the image is also straight, meaning the image is erect.

Step 2: The image height is 25 cm, while the object height is 100 cm. This means the image is diminished.

Step 3: A diminished and erect image is always formed by a convex mirror.

Step 4: Convex mirrors always form virtual images on the opposite side of the mirror.

Why other options are wrong:
- Option A: Concave mirrors can form real images, but they are inverted. Also, the problem states the image is erect.
- Option B: Concave mirrors can form virtual images, but they are magnified when the object is between the pole and the focus. Here, the image is diminished.
- Option C: Convex mirrors always form virtual images, not real images.

Key Concepts:
- Image formation by convex and concave mirrors
- Real and virtual images
- Magnification

Memory Aids:
- Remember the characteristics of images formed by different types of mirrors.

Quick Tips:
- Know the image formation rules for different types of mirrors.

17 Medium

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Application +4 | -1

In an experiment a convex lens of focal length 15 cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5 cm from it. It is found that an object and its image coincide, if the object is placed at a distance of 20 cm from the lens. The focal length of the convex mirror is:

View answer and explanation

Correct option: A

27.5 cm

Option A (27.5 cm) is correct because the object and image coincide when the ray retraces its path, indicating the object is at the center of curvature of the mirror.

Solution approach

Correct Answer + Core Reason: 27.5 cm. The object and image coincide when the ray retraces its path, meaning the object is at the center of curvature of the mirror.

Step-by-Step Solution:

Step 1: Identify given information: Focal length of convex lens, ${f_L} = 15 \text{{ cm}}$ , distance between lens and mirror, $d = 5 \text{{ cm}}$ , object distance from lens, $u = -20 \text{{ cm}}$ .

Step 2: Find the image distance $v$ for the lens using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ .

Step 3: Substitute the values: $\frac{1}{{15}} = \frac{1}{v} - \frac{1}{{-20}}$ , so $\frac{1}{v} = \frac{1}{{15}} - \frac{1}{{20}} = \frac{{4 - 3}}{{60}} = \frac{1}{{60}}$ .

Step 4: Therefore, $v = 60 \text{{ cm}}$ .

Step 5: The distance of the image formed by the lens from the mirror is $60 - 5 = 55 \text{{ cm}}$ .

Step 6: Since the object and image coincide, the image formed by the lens acts as an object for the mirror, and the final image is formed at the same location as the object. This means the image formed by the lens is at the center of curvature of the mirror.

Step 7: Therefore, the radius of curvature of the mirror is $R = 55 \text{{ cm}}$ .

Step 8: The focal length of the convex mirror is $f = \frac{R}{2} = \frac{{55}}{2} = 27.5 \text{{ cm}}$ .

Why other options are wrong:
- Option B (20.0 cm): This is incorrect. It might be obtained by incorrectly assuming the object distance from the lens is the radius of curvature of the mirror.
- Option C (25.0 cm): This is incorrect. It's a value close to the correct answer but obtained through incorrect calculations.
- Option D (30.5 cm): This is incorrect. It's a value close to the correct answer but obtained through incorrect calculations.

Key Concepts:
- Lens formula
- Mirror formula
- Radius of curvature and focal length relationship

Memory Aids:
- Remember the sign conventions for lenses and mirrors.

Quick Tips:
- When object and image coincide, the rays are retracing their path.
- The image formed by the lens acts as an object for the mirror.

18 Medium

18 of 241Question permalink

Application +4 | -1

. An movie projector forms an image

3.5\,\,m

long of an object

35\,\,mm

. Supposing there is negligible absorption of light by aperture then illuminance on slide and screen will be in the ratio of

View answer and explanation

Correct option: B

{10^4}:1

Option B (

{10^4}:1

) is correct because the illuminance is inversely proportional to the square of the image size (or the square of the magnification).

Solution approach

Correct Answer + Core Reason: ${10^4}:1$ . The illuminance is inversely proportional to the square of the magnification.

Step-by-Step Solution:

Step 1: Define illuminance.

Step 2: Key Concept: Illuminance (E) is the luminous flux per unit area. It is inversely proportional to the square of the distance from the source.

Step 3: Key Concept: The ratio of illuminance on the slide to the screen is inversely proportional to the square of the magnification (m).
$\frac{E_{slide}}{E_{screen}} = m^2$

Step 4: Calculate the magnification: $m = \frac{Image \, size}{Object \, size} = \frac{3.5 \, m}{35 \, mm} = \frac{3.5 \, m}{0.035 \, m} = 100$

Step 5: Calculate the ratio of illuminance: $\frac{E_{slide}}{E_{screen}} = (100)^2 = 10000$

Step 6: Therefore, the ratio is $10000:1$ or ${10^4}:1$ .

Why other options are wrong:
- Option A: $100:1$ . This is the magnification, not the square of the magnification.
- Option C: $1:100$ . This is the inverse of the magnification.
- Option D: $1:{10^4}$ . This is the inverse of the square of the magnification.

Key Concepts:
- Illuminance
- Magnification

Memory Aids:
- Illuminance is inversely proportional to the square of the distance.

Quick Tips:
- Remember the relationship between illuminance and magnification.

19 Medium

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Analyzing +4 | -1

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: The phase difference of two light waves change if they travel through different media having same thickness, but different indices of refraction.

Reason R: The wavelengths of waves are different in different media.

In the light of the above statements, choose the most appropriate answer from the options given below

View answer and explanation

Correct option: D

Both A and R are correct and R is the correct explanation of A

Option D (Both A and R are correct and R is the correct explanation of A) is correct because the phase difference depends on the wavelength, which changes in different media.

Solution approach

Correct Answer + Core Reason: Both A and R are correct and R is the correct explanation of A. The phase difference changes due to the change in wavelength in different media.

Step-by-Step Solution:

Step 1: The phase difference $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$ , where $\lambda$ is the wavelength and $\Delta x$ is the path difference.

Step 2: When light travels through different media with the same thickness, the path difference is the same, but the wavelength changes because the refractive index is different.

Step 3: Since the wavelength changes, the phase difference also changes.

Step 4: Therefore, both the assertion and the reason are correct, and the reason is the correct explanation of the assertion.

Why other options are wrong:
- Option A: The reason is the correct explanation of the assertion.
- Option B: The assertion is correct.
- Option C: The reason is the correct explanation of the assertion.

Key Concepts:
- Phase difference
- Wavelength
- Refractive index

Memory Aids:
- Remember the relationship between phase difference, wavelength, and path difference.

Quick Tips:
- Understand how the refractive index affects the wavelength of light.

20 Easy

20 of 241Question permalink

Understanding +4 | -1

The passenger side-view mirror on an automobile often has the notation. "Objects seen in mirror are closer than they appear". Is the image really farther away than the object?

View answer and explanation

Correct option: B

No, the image is smaller and closer than the object

Option B is correct because the convex mirror used in the side-view mirror creates a smaller, upright, and virtual image that appears closer than the actual object.

Solution approach

Correct Answer + Core Reason: No, the image is smaller and closer than the object. Convex mirrors produce smaller, closer images to provide a wider field of view.

Step-by-Step Solution:

Step 1: Understand the properties of convex mirrors.

Step 2: Convex mirrors always produce virtual, upright, and diminished (smaller) images.

Step 3: Because the image is smaller, the brain perceives it as being closer than it actually is. This is why the warning is present.

Step 4: Therefore, the image is not farther away; it is closer and smaller.

Why other options are wrong:
- Option A: This is incorrect because while the image is smaller, it is not farther away. The warning indicates the opposite.
- Option C: This is incorrect because the image is not larger; it is smaller. Convex mirrors always produce diminished images.
- Option D: This is incorrect because the image is not larger and farther away. The image is smaller and closer.

Key Concepts:
- Convex mirrors
- Image formation by mirrors
- Properties of images (real/virtual, upright/inverted, magnified/diminished)

Memory Aids:
- Convex mirrors: "Smaller and closer"

Quick Tips:
- Remember that convex mirrors always produce virtual, upright, and smaller images. This is why they are used as side-view mirrors to provide a wider field of view.

21 Medium

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Understanding +4 | -1

The aperture of a telescope is made large, because

View answer and explanation

Correct option: A

To increase the intensity of image

Option A (To increase the intensity of image) is correct because a larger aperture allows more light to enter the telescope, resulting in a brighter and more intense image.

Solution approach

Correct Answer + Core Reason: To increase the intensity of image. A larger aperture gathers more light.

Step-by-Step Solution:

Step 1: Understand the function of the aperture.\nThe aperture of a telescope is the diameter of its objective lens or mirror. The larger the aperture, the more light it can gather from distant objects.

Step 2: Relate aperture to image intensity.\nThe intensity of the image is directly proportional to the amount of light gathered by the telescope. Therefore, a larger aperture results in a brighter and more intense image.

Step 3: Final Answer: To increase the intensity of image

Why other options are wrong:
- Option B (To decrease the intensity of image): This is incorrect because a larger aperture increases the intensity of the image.
- Option C (To have greater magnification): While a larger aperture can improve the quality of the image, it does not directly increase the magnification. Magnification is determined by the focal lengths of the objective and eyepiece.
- Option D (To have lesser resolution): This is incorrect because a larger aperture increases the resolution of the telescope.

Key Concepts:
- Telescope
- Aperture
- Intensity of image
- Resolution

Memory Aids:
- Bigger aperture = Brighter image.

Quick Tips:
- Remember that a larger aperture increases the amount of light gathered, leading to a brighter image.

22 Easy

22 of 241Question permalink

Application +4 | -1

Light of wavelength

500 \mathrm{~nm}

travelling with a speed of

2.0 \times 10^8 \mathrm{~ms}^{-1}

in a certain medium enters another medium of refractive index

5 / 4

times that of the first medium. What are the wavelength and speed in the second medium?

View answer and explanation

Correct option: A

Wavelength/nm - 400, Speed

/\mathrm{ms}^{-1}

-

1.6 \times 10^8

Option A (Wavelength/nm - 400, Speed

/\mathrm{ms}^{-1}

-

1.6 \times 10^8

) is correct because both wavelength and speed decrease when light enters a denser medium.

Solution approach

Correct Answer + Core Reason: Wavelength/nm - 400, Speed $/\mathrm{ms}^{-1}$ - $1.6 \times 10^8$ . Both wavelength and speed decrease in a denser medium.

Step-by-Step Solution:

Step 1: Identify given information: Initial wavelength $\lambda_1 = 500 \mathrm{~nm}$ , initial speed $v_1 = 2.0 \times 10^8 \mathrm{~ms}^{-1}$ , refractive index ratio $n_2/n_1 = 5/4$ .

Step 2: Recall the relationship between refractive index, speed, and wavelength: $n = c/v$ and $\lambda = v/f$ , where $f$ is the frequency (which remains constant when light enters a new medium).\nTherefore, $n_1 = c/v_1$ and $n_2 = c/v_2$ , and $n_2/n_1 = v_1/v_2$ .\nAlso, $\lambda_1 = v_1/f$ and $\lambda_2 = v_2/f$ , so $\lambda_2/\lambda_1 = v_2/v_1$ .

Step 3: Calculate the speed in the second medium: $v_2 = v_1 \times (n_1/n_2) = (2.0 \times 10^8 \mathrm{~ms}^{-1}) \times (4/5) = 1.6 \times 10^8 \mathrm{~ms}^{-1}$ .

Step 4: Calculate the wavelength in the second medium: $\lambda_2 = \lambda_1 \times (v_2/v_1) = (500 \mathrm{~nm}) \times (1.6 \times 10^8 \mathrm{~ms}^{-1}) / (2.0 \times 10^8 \mathrm{~ms}^{-1}) = 500 \mathrm{~nm} \times (4/5) = 400 \mathrm{~nm}$ .

Step 5: Final Answer: Wavelength = 400 nm, Speed = $1.6 \times 10^8 \mathrm{~ms}^{-1}$ .

Why other options are wrong:
- Option B: This option has the correct wavelength but an incorrect speed. The speed should decrease, not increase.
- Option C: This option has the initial wavelength, which is incorrect. The wavelength changes when the medium changes.
- Option D: This option has an incorrect wavelength and speed. The wavelength should decrease, and the speed should also decrease.

Key Concepts:
- Refractive index and its relation to speed of light: $n = c/v$
- Wavelength and its relation to speed and frequency: $\lambda = v/f$
- Frequency remains constant when light travels from one medium to another.

Memory Aids:
- Remember that speed and wavelength are inversely proportional to the refractive index.

Quick Tips:
- Always remember that the frequency of light remains constant when it travels from one medium to another. Only speed and wavelength change.

23 Medium

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Remembering +4 | -1

The spectrum obtained from an electric lamp or red hot heater is

View answer and explanation

Correct option: D

Continuous spectrum

Option D (Continuous spectrum) is correct because an electric lamp or red hot heater emits light of all wavelengths.

Solution approach

Correct Answer + Core Reason: Continuous spectrum. An electric lamp or red hot heater emits light of all wavelengths, producing a continuous spectrum.

Step-by-Step Solution:

Step 1: A continuous spectrum contains all wavelengths of light.

Step 2: An electric lamp or red hot heater emits light due to thermal radiation.

Step 3: Thermal radiation produces light of all wavelengths, resulting in a continuous spectrum.

Why other options are wrong:
- Option A: Line spectrum. This is incorrect because a line spectrum consists of discrete lines at specific wavelengths.
- Option B: Band spectrum. This is incorrect because a band spectrum is produced by molecules and consists of bands of closely spaced lines.
- Option C: Absorption spectrum. This is incorrect because an absorption spectrum is produced when light passes through a substance, and certain wavelengths are absorbed.

Key Concepts:
- Continuous spectrum
- Line spectrum
- Band spectrum
- Absorption spectrum
- Thermal radiation

Memory Aids:
- Remember that a continuous spectrum contains all wavelengths of light.

Quick Tips:
- Understand the different types of spectra and their origins.

24 Easy

24 of 241Question permalink

Application +4 | -1

Critical angle of incidence for a pair of optical media is

45^{\circ}

. The refractive indices of first and second media are in the ratio:

View answer and explanation

Correct option: A

\sqrt{2}: 1

Option A (

\sqrt{2}: 1

) is correct because the ratio of refractive indices is inversely proportional to the sine of the critical angle.

Solution approach

Correct Answer + Core Reason: $\sqrt{2}: 1$ . This is because the critical angle is related to the ratio of refractive indices.

Step-by-Step Solution:

Step 1: Identify given information: Critical angle $C = 45^{\circ}$ .

Step 2: Recall the definition of critical angle: $\sin C = n_2/n_1$ , where $n_1$ is the refractive index of the denser medium and $n_2$ is the refractive index of the rarer medium.

Step 3: Substitute the given value: $\sin 45^{\circ} = n_2/n_1$ .

Step 4: Evaluate $\sin 45^{\circ}$ : $\sin 45^{\circ} = 1/\sqrt{2}$ .

Step 5: Find the ratio $n_1/n_2$ : $n_1/n_2 = \sqrt{2}/1$ .

Step 6: Final Answer: The ratio of refractive indices is $\sqrt{2}: 1$ .

Why other options are wrong:
- Option B: $1: 2$ is incorrect because it's the inverse of the correct ratio squared.
- Option C: $1: \sqrt{2}$ is incorrect because it's the inverse of the correct ratio.
- Option D: $2: 1$ is incorrect because it's the square of the correct ratio.

Key Concepts:
- Critical angle: $\sin C = n_2/n_1$

Memory Aids:
- Remember that the critical angle occurs when light travels from a denser to a rarer medium.

Quick Tips:
- Always remember the relationship between critical angle and refractive indices.

25 Easy

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Analyzing +4 | -1

Mimimum distance between object and image formed by concave mirror of focal length

f

can be:

View answer and explanation

Correct option: D

zero

Option D (zero) is correct because the minimum distance between an object and its image formed by a concave mirror is zero when the object is at the focal point.

Solution approach

Correct Answer + Core Reason: zero. The minimum distance between the object and image formed by a concave mirror can be zero when the object is placed at the center of curvature.

Step-by-Step Solution:

Step 1: Consider the object placed at the center of curvature (2f) of the concave mirror. In this case, the image is also formed at the center of curvature (2f), and it is real and inverted.

Step 2: The distance between the object and the image is zero. If the object is placed at the focus, the image is formed at infinity. If the object is placed at infinity, the image is formed at the focus.

Step 3: When the object is at the center of curvature (C), the image is also at C. Therefore, the minimum distance between the object and the image is zero.

Why other options are wrong:
- Option A: This is incorrect because the distance between the object and image can be zero.
- Option B: This is incorrect because the distance between the object and image can be zero.
- Option C: This is incorrect because the distance between the object and image can be zero.

Key Concepts:
- Concave mirror
- Image formation
- Center of curvature
- Focal length

Memory Aids:
- Object at C, image at C.

Quick Tips:
- Remember the special cases for image formation in concave mirrors.

26 Medium

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Analyzing +4 | -1

Assertion: In ionospheric reflection, phase change does not occurs with the light wave.
Reason: The ionosphere reflection is similar to the total internal reflection which is similar to the total internal reflection in miraj

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion

Option A (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion) is correct because ionospheric reflection is similar to total internal reflection, and no phase change occurs.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. Ionospheric reflection is similar to total internal reflection, and no phase change occurs.

Step-by-Step Solution:

Step 1: Understand Ionospheric Reflection: In ionospheric reflection, radio waves are bent back towards the Earth due to the varying refractive index of the ionosphere. This bending is similar to total internal reflection.

Step 2: Understand Phase Change: During total internal reflection, there is no phase change of the wave. This is because the wave is completely reflected, not transmitted.

Step 3: Analyze the Assertion: The assertion states that no phase change occurs during ionospheric reflection, which is correct.

Step 4: Analyze the Reason: The reason states that ionospheric reflection is similar to total internal reflection, which is also correct. The reason correctly explains why there is no phase change.

Why other options are wrong:
- Option B: The reason is the correct explanation of the assertion.
- Option C: Both assertion and reason are correct.
- Option D: Both assertion and reason are correct.

Key Concepts:
- Ionospheric Reflection
- Total Internal Reflection
- Phase Change

Memory Aids:
- Remember that total internal reflection does not involve a phase change.

Quick Tips:
- Understand the similarities between ionospheric reflection and total internal reflection.

27 Medium

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Application +4 | -1

A person cannot see properly beyond

\mathrm{2}\,\,\mathrm{m}

. Power of the lens is

View answer and explanation

Correct option: D

\mathrm{ - 0}\mathrm{.5}\,\,\mathrm{D}

Option D (

\mathrm{ - 0}\mathrm{.5}\,\mathrm{D}

) is correct because a person who cannot see beyond 2m needs a diverging lens with power -0.5 D.

Solution approach

Correct Answer + Core Reason: $\mathrm{ - 0}\mathrm{.5}\,\mathrm{D}$ . The power of the lens is calculated using the formula for correcting myopia.

Step-by-Step Solution:

Step 1: The person cannot see properly beyond 2 m, which means the far point is 2 m. We need to find the power of the lens required to correct this.

Step 2: The object distance is infinity ( $u = \infty$ ) and the image distance should be 2 m ( $v = -2 \, \mathrm{m}$ ) for the person to see clearly.

Step 3: Using the lens formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ , we have $\frac{1}{f} = \frac{1}{-2} - \frac{1}{\infty} = -\frac{1}{2}$ .

Step 4: Therefore, the focal length $f = -2 \, \mathrm{m}$ .

Step 5: The power of the lens is given by $P = \frac{1}{f}$ (in meters). So, $P = \frac{1}{-2} = -0.5 \, \mathrm{D}$ .

Why other options are wrong:
- Option A: This is incorrect because it is the reciprocal of the far point distance but with the wrong sign. It would be the power of a converging lens.
- Option B: This is incorrect because it is not related to the given far point distance.
- Option C: This is incorrect because it is not related to the given far point distance and has the wrong sign.

Key Concepts:
- Myopia (nearsightedness)
- Lens formula
- Power of a lens

Memory Aids:
- For myopia, a diverging lens is used (negative power).
- Power (in diopters) = 1 / focal length (in meters)

Quick Tips:
- Remember that for myopia, the image is formed in front of the retina, so a diverging lens is needed to correct it.
- Use the correct sign conventions for object and image distances.

28 Easy

28 of 241Question permalink

Application +4 | -1

Light enters from air into a given medium at an angle of

45^{\circ}

with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of

15^{\circ}

from its original direction. The refractive index of the medium is

View answer and explanation

Correct option: C

1.414

Option C (

1.414

) is correct because it is calculated using Snell's law, considering the deviation angle.

Solution approach

Correct Answer + Core Reason: $1.414$ . The refractive index is calculated using Snell's Law and the given deviation angle.

Step-by-Step Solution:

Step 1: The angle of incidence is given as $i = 45^{\circ}$ .

Step 2: The angle of deviation is given as $\delta = 15^{\circ}$ .

Step 3: The angle of refraction $r$ can be found using the relationship $i - r = \delta$ , so $r = i - \delta = 45^{\circ} - 15^{\circ} = 30^{\circ}$ .

Step 4: Snell's law states that $n_1 \sin i = n_2 \sin r$ , where $n_1$ is the refractive index of air (approximately 1) and $n_2$ is the refractive index of the medium.

Step 5: Therefore, $1 \times \sin 45^{\circ} = n_2 \times \sin 30^{\circ}$ .

Step 6: $n_2 = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414$ .

Why other options are wrong:
- Option A: This value is incorrect and doesn't result from applying Snell's law correctly with the given angles.
- Option B: This value is incorrect and doesn't result from applying Snell's law correctly with the given angles.
- Option D: This value is incorrect and doesn't result from applying Snell's law correctly with the given angles.

Key Concepts:
- Snell's Law
- Refractive Index
- Angle of Incidence
- Angle of Refraction
- Angle of Deviation

Memory Aids:
- Snell's Law: $n_1 \sin i = n_2 \sin r$ .

Quick Tips:
- Always remember to calculate the angle of refraction correctly using the deviation angle.

29 Medium

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Application +4 | -1

Find the luminous intensity of the sun if it produces the same illuminance on the earth as produced by a bulb of

10000\,\,candela

at a distance

0.3\,\,m

. The distance between the sun and the earth is

1.5\,\, \times \,\,{10^{11}}\,\,m

View answer and explanation

Correct option: C

25\,\, \times \,\,{10^{26}}\,\,cd

Option C (

25 \times {10^{26}}\,cd

) is correct because it is calculated by equating the illuminance produced by the sun and the bulb and then solving for the luminous intensity of the sun.

Solution approach

Correct Answer + Core Reason: $25 \times {10^{26}}\,cd$ . The luminous intensity is calculated by equating the illuminance produced by the sun and the bulb.

Step-by-Step Solution:

Step 1: Define the given quantities.\nIlluminance produced by the bulb = Illuminance produced by the sun\nLuminous intensity of the bulb, $I_b = 10000 \, candela$ \nDistance of the bulb, $d_b = 0.3 \, m$ \nDistance of the sun, $d_s = 1.5 \times 10^{11} \, m$ \nLet the luminous intensity of the sun be $I_s$ .

Step 2: Write the formula for illuminance.\nIlluminance, $E = \frac{I}{d^2}$ , where $I$ is the luminous intensity and $d$ is the distance.

Step 3: Equate the illuminance produced by the bulb and the sun.
$\frac{I_b}{d_b^2} = \frac{I_s}{d_s^2}$

Step 4: Substitute the given values.
$\frac{10000}{(0.3)^2} = \frac{I_s}{(1.5 \times 10^{11})^2}$

Step 5: Solve for $I_s$ .
$I_s = \frac{10000 \times (1.5 \times 10^{11})^2}{(0.3)^2}$
$I_s = \frac{10^4 \times 2.25 \times 10^{22}}{0.09}$
$I_s = \frac{2.25 \times 10^{26}}{0.09}$
$I_s = 25 \times 10^{26} \, candela$

Why other options are wrong:
- Option A: $25 \times {10^{22}}\,cd$ This is incorrect because the calculation of the luminous intensity involves squaring the distance, which is not done correctly in this case.
- Option B: $25 \times {10^{18}}\,cd$ This is incorrect because the powers of 10 are not handled correctly during the calculation.
- Option D: $25 \times {10^{36}}\,cd$ This is incorrect because the powers of 10 are not handled correctly during the calculation. The exponent should be 26, not 36.

Key Concepts:
- Illuminance
- Luminous intensity
- Inverse square law

Memory Aids:
- Remember the formula for illuminance: $E = \frac{I}{d^2}$

Quick Tips:
- Make sure to use consistent units.
- Pay attention to the powers of 10 in the calculations.

30 Medium

30 of 241Question permalink

Understanding +4 | -1

Relation between critical angles of water and glass is

View answer and explanation

Correct option: A

{C_\omega } > {C_g}

Option A (

{C_\omega } > {C_g}

) is correct because water has a lower refractive index than glass, resulting in a larger critical angle for water.

Solution approach

Correct Answer + Core Reason: ${C_\omega } > {C_g}$ . Water has a lower refractive index than glass, leading to a larger critical angle.

Step-by-Step Solution:

Step 1: Critical Angle Formula: The critical angle ( $C$ ) is given by $\sin C = \frac{1}{n}$ , where $n$ is the refractive index of the denser medium with respect to the rarer medium (usually air).

Step 2: Refractive Indices: The refractive index of water ( $n_\omega$ ) is approximately 1.33, and the refractive index of glass ( $n_g$ ) is approximately 1.5.

Step 3: Compare Critical Angles: Since $n_\omega < n_g$ , then $\frac{1}{n_\omega} > \frac{1}{n_g}$ . Therefore, $\sin C_\omega > \sin C_g$ , which implies $C_\omega > C_g$ .

Why other options are wrong:
- Option B: ${C_\omega } < {C_g}$ : Incorrect because water has a lower refractive index than glass.
- Option C: ${C_\omega } = {C_g}$ : Incorrect because the refractive indices are different.
- Option D: ${C_\omega } = {C_g} = 0$ : Incorrect because critical angles are not zero.

Key Concepts:
- Critical angle
- Refractive index
- Total Internal Reflection

Memory Aids:
- Lower refractive index → Larger critical angle.

Quick Tips:
- Remember the relationship between refractive index and critical angle.

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