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}=\frac{1}{3}-\frac{1}{9}=\frac{2}{9} $
Therefore $ v _{1}=\frac{3}{2}v $
For surface Q, $ \frac{1}{v _{2}}=\frac{1}{f}-\frac{1}{u}=\frac{1}{5}-\frac{1}{15}=\frac{2}{15} $
Therefore $ v _{2}=\frac{5}{4}m $
$ v _{1}-v _{2}=0.25\ \text{m} $
Magnification of $ P=\frac{v _{1}}{u}=\frac{3/2}{3}=\frac{1}{2} $
Height of $ P=\frac{1}{2}\times 2=1\text{ m} $
Magnification of $ Q=\frac{v}{u}=\frac{5/4}{5}=\frac{1}{4} $
Height of $ Q=\frac{1}{4}\times 2=0.5m $
BETA
