Question

A needle lying on a horizontal surface is viewed vertically from a distance of 100 cm. A glass slab of 12 cm thickness is placed on the needle. Find the distance of the vertical image of the needle from the eye, taking _an_g = 1.5.

Answer 1

Given: Actual distance from eye = 100 cm

Thickness of glass slab = 12 cm

Refractive index of glass, n = 1.5

The slab reduces the apparent depth of the needle. The apparent thickness of the slab is given by:

Apparent thickness = `"Actual thickness"/n`

= `12/1.5`

= 8 cm

Now,

Distance of image from eye = 100 − 12 + 8 = 96 cm

The vertical image of the needle appears at a distance of 96 cm from the eye.