Question

In displacement method, the distance between object and screen is 96cm. The ratio of length of two images formed by a convex lens placed between them is 4.84.

• A. Ratio of the length of object to the length of shorter image is $\frac{11}{5}$
• B. Distance between the two positions of the lens is 36cm.
• C. Focal length of the lens is 22.5cm.
• D. Distance of the lens from the shorter image is 30cm.

Answer 1

Answer: (A), (B), (D)

The correct options are
A Ratio of the length of object to the length of shorter image is $\frac{11}{5}$
B Distance between the two positions of the lens is 36cm.
D Distance of the lens from the shorter image is 30cm.

D = 96
Given $\frac{I_2}{I_1}=4.84$
Let $I_2$ = 4.84 a and $I_1$ = a
$O=\sqrt{I_1{I}_2}=2.2a$
(A) Required ratio $=\frac{2.2a}{a}=\frac{11}{5}$
(B) $\frac{v}{u}=\frac{11}{5}$ . . . (1)
v + u = 96 ….(2)
$\therefore v+\frac{5v}{11}=96\Rightarrow \frac{16v}{11}=96$
$\Rightarrow$ v = 66 and u = 30
Distance between two positions
= v – u = 66 – 30 = 36 cm
(C) $\frac{1}{f}=\frac{1}{66}+\frac{1}{30}=\frac{30+66}{30\times 66}\Rightarrow f=\frac{30\times 66}{96}=\frac{330}{16}=20.625cm$
(D) v = 30 for shorter image.