Question

In displacement method, the distance between object and screen is $96\text{cm}$. The ratio of length of two images formed by a convex lens placed between them is $4.84$. Then choose the correct option(s).

• A. The ratio of the length of object to the length of shorter image is $11/5$.
• B. The distance between the two positions of the lens is $36\text{cm}$.
• C. The focal length of the lens is $22.5\text{cm}$.
• D. The distance of the lens from the shorter image is $30\text{cm}$.

Answer 1

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

The distance of the lens from the shorter image is $30\text{cm}$.


Given:

$D=96\text{cm};\frac{h_2}{h_1}=4.84$

From displacement method,

$h_o=\sqrt{h_1{h}_2}...\left(1\right)$, and

$f=\frac{D^2-d^2}{4D}...\left(2\right)$

Now,

Ratio of the length of the object to the length of the shorter image,

$\frac{h_o}{h_1}=\sqrt{\frac{h_2}{h_1}}\left[\text{From (1)}\right]$

$\frac{h_o}{h_1}=\sqrt{4.84}=\frac{11}{5}$

Now, $m=\frac{v}{u}=\frac{h_1}{h_o}=\frac{5}{11}$

$\Rightarrow u=2.2v\Rightarrow x_1=2.2y_1$

Also, $x_1+y_1=D\Rightarrow 2.2y_1+y_1=96$

$\Rightarrow y_1=30\text{cm}\left[\text{Distance of the lens from the shorter image}\right]$

And $x_1=2.2y_1=2.2\times 30=66\text{cm}$

The distance between the two positions of the lens,

$d=x_1-x_2=x_1-y_1=66-30=36\text{cm}[\text{From symmetry,}{x}_2=y_1]$

Further, focal length,

$f=\frac{{96}^2-{36}^2}{4\times 96}=20.6\text{cm}\left[\text{From (2)}\right]$