Optics Question 120

Question: A cube of side 2 m is placed in front of a concave mirror focal length 1m with its face P at a distance of 3 m and face Q at a distance of 5 m from the mirror. The distance between the images of face P and Q and height of images of P and Q are

Options:

A) 1 m, 0.5 m, 0.25 m

B) 0.5 m, 1 m, 0.25 m

C) 0.5 m, 0.25 m, 1m

D) 0.25 m, 1m, 0.5 m

Show Answer

Answer:

Correct Answer: D

Solution:

For surface P, $ \frac{1}{v _{1}}=\frac{1}{f}-\frac{1}{u}=1-\frac{1}{3}=\frac{2}{3} $

Therefore $ v _{1}=\frac{3}{2}m $

For surface Q, $ \frac{1}{v _{2}}=\frac{1}{f}-\frac{1}{u}=1-\frac{1}{5}=\frac{4}{5} $

Therefore $ v _{2}=\frac{5}{4}m $

$ v _{1}-v _{2}=0.25m $

Magnification of $ P=\frac{v _{1}}{u}=\frac{3/2}{3}=\frac{1}{2} $

Height of $ P=\frac{1}{2}\times 2=1m $

Magnification of $ Q=\frac{v _{2}}{u}=\frac{5/4}{5}=\frac{1}{4} $

Height of $ Q=\frac{1}{4}\times 2=0.5m $



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