Question

The lens is an overhead projector forms an image P' of a point P on an overhead transparency. If the screen is moved closer to the projector to keep the image on the screen in focus, the lens must be

• A. moved up
• B. moved towards left
• C. moved down
• D. moved lens of focal

Answer 1

The correct option is A moved up

First the image is at P' and the lens is positioned at $L_1$.

The final image can be considered to have formed without reflection from the mirror.

So, $\frac{1}{v_0}=\frac{1}{f}+\frac{1}{u_0}$

Now the screen is moved towards the mirror which means that the final image is formed at a lesser distance by x, as shown in figure.

Let the distance lens need to be moved be y.

We need to find the sign of this y.

From figure, considering refraction from lens at $L_2$,

$\frac{1}{v_0-x}=\frac{1}{u_0-y}+\frac{1}{f}$

Since x is positive,

Thus , $\frac{1}{u_0-y}+\frac{1}{f}>\frac{1}{v_0}$

Hence, $\frac{1}{u_0-y}>\frac{1}{u_0}$

Hence y is positive, that means lens needs to be moved up.