Question

A lens with a focal length of $f=30\text{cm}$ produces on a screen a sharp image of an object that is at a distance of $a=40\text{cm}$ from the lens. A plane parallel plate with thickness of $d=9\text{cm}$ is placed between the lens and the object perpendicular to the optical axis of the lens. Through what distance (in $\text{cm})$ should the screen be shifted for the image of the object to remain distinct? The refractive index of the glass of the plate is $\mu =1.8$

Answer 1

Using lens formula , $\frac{1}{v}-\frac{1}{u}-\frac{1}{f}$

$\frac{1}{v_1}-\frac{1}{-40}=\frac{1}{30}$

$\therefore v_1=120\text{cm}$

Shift due to the slab,
$\Delta x=(1-\frac{1}{\mu })d=(1-\frac{1}{1.8})9=4\text{cm}$

$\therefore u^{\prime }=-(40-\Delta x)=-36\text{cm}$

Thus, $\frac{1}{v_2}-\frac{1}{-36}=\frac{1}{30}$

$\Rightarrow v_2=180\text{cm}$

Therefore screen needs to be shifted by a distance $|v_2-v_1|=|180-120|=60\text{cm}$ away from the lens

Accepted answers: $60,60.0,60.00$