Answer to Question #88268 in Optics for .xjghdxf

Question

An object is placed 2.4m from a convex lens. It forms an image 12cm from the lens on a screen. A glass slab of thickness 1cm and refractive index 1.5 is introduced between the lens and screen. By what distance should the object be moved so that the image is in focus?

Answer 1

For case I (without a glass slab)

Formula of convex lens:

$$\frac {1}{d} + \frac {1}{d_i} = \frac {1}{f} (1)$$

In our case, d=240 cm, d_i=12 cm

Using (1) we got:

focal length of the lens f=11.43 cm

For case II (with a glass slab)

We can write for a glass slab:

$$shift = 1 - \frac {2}{3} = 0.33 (2)$$

Distance from the lens to the image:

d_i=12-0.33=11.67 cm

Formula of convex lens:

$$\frac {1}{d} + \frac {1}{d_i} = \frac {1}{f} (3)$$

In our case, d_i=11.67 cm, f=11.43 cm

Using (3) we got:

d=5.56 m

Distance:

Δd=5.56 m-2.4 m=3.16 m

Answer:

3.16 m