Question

In displacement method , the distance between object and screen is $96cm$ . The ration 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 $11/5$
• B. Distance between the two position 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)

Distance between the two position of the lens is $36cm$
Ratio of the length of object to the length of shorter image is $11/5$
Distance of the lens from the shorter image is $30cm$

Given $d=96$ $cm$, ratio= $4.84$

lets write larger image = $I_1$, shorter image = $I_2$ and size of object $=O$

according to question $I_1/I_2=4.84$ i.e $I_1=4.84I_2$

we know magnification $m=\frac{v}{u}$

for first position $m=\frac{v}{u}=\frac{I_1}{O}$

for second position $m=\frac{u}{v}=\frac{I_2}{O}$

$⟹\frac{v}{u}=\frac{I_1}{O}=\frac{O}{I_2}$

O = $\sqrt[2]{2I_1.I_2}$

O = $\sqrt[2]{2I_1.I_2}$= $\sqrt[2]{2I_2.4.84I_2}$

= $\sqrt[2]{24.84I_2^2}$ = $2.2I_2$

=$\frac{11}{5}{I}_2$

which is Ratio of the length of object to the length of shorter image and that is is 11/5, option A is correct.

$\frac{v}{u}=\frac{O}{I_2}=2.2$ and given $v+u=96$

$2.2u+u=96⟹u=30$ and $v=66$

Distance between two position of the lens = $66-30=36$ $cm$

and shorter image distance $u=30$ $cm$

option B and D are also correct

to calculate focal length $\frac{1}{f}$=$\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f}$=$\frac{1}{66}-\frac{1}{-30}$

$\frac{1}{f}$=$\frac{30+66}{66.30}$

$⟹f=20.6$ $cm$ option C is not true.

Option A,B and D are correct.