Back to RAY OPTICS

NEET UG PHYSICS MCQ: RAY OPTICS

1 Hard

1 of 1

Application +4 | -1

When the distance between the object and the screen is more than

4 \mathrm{~F}

, we can obtain image of an object on the screen for the two positions of a lens. It is called displacement method. In one case, the image is magnified and in the other case it is diminished. Then, the ratio of the size of image to the diminished image is

View answer and explanation

Correct option: A

\frac{(D+d)^2}{(D-d)^2}

The correct answer is A (

\frac{{(D+d)^2}}{{(D-d)^2}}

) because it represents the ratio of image sizes in the displacement method, derived from the lens formula and magnification.

More Questions from RAY OPTICS

Practice All Chapter MCQs →

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 Red colour is used for danger signals because 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 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. 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. If both the object and image are at infinite distance from a refracting telescope its magnifying power will be equal to