Question

Question

An ink dot on the sheet of a paper is viewed from above a distance of 25 cm. By what distance would the ink dot appear to be raised if it is viewed through a glass slab of thickness 10 cm held // to the paper? Given µglass= 1.5

answer should be proper otherwise it will get reported

Answer 1

Given: An ink dot on the sheet of a paper is viewed from above at a distance of 25cm. A glass slab of thickness 10 cm is held parallel to the paper. Refractive index of glass= 1.5

To find: By what distance would the ink dot appear to be raised

Explanation: Refractive index= 1.5

When a glass slab is kept parallel to the paper, the ray of light undergoes refraction and the object appears to be raised, that is, seems closer than it actually is.

Formula for calculating the extra distance is given by: t(1-1/u) where t is the thickness of the slab and u is the refractive index of the glass.

Now, 1-1/u = 1 - 1/1.5

= 1.5-1 / 1.5

= 1/3

t= 10 cm

Using formula given above: Shift= 10* 1/3

= 3.33 cm

Therefore, the ink dot appears to be raised by 3.33 cm.