Question

A glass sphere ( n= 1.5 ) 4 CM in diameter has a Luminous speck O embedded inside the sphere 1 cm below the surface find where it will appear to be when vied along adiametre of the sphere from the near surface

Answer 1

Given:

The refractive index of glass n= 1.5

The refractive index of air = 1

Diameter d=4 cm

Inside length =1 cm

To find:

Distance of the where it appear.

Solution:

Object distance from the sphere u=4-1= 3 cm

Radius of the sphere R=4/2=2 cm

Using the refractive index formula

n(air)/v-n/u=(n(air) - n)/R

1/v -1.5/3= (1-1.5)/2

1/v=0.5/2 + 1.5/3

1/v=3+3/6

1/v=1

v=1 cm

So, the distance is appear is v=1 cm

Answer 2

The distance of the image is 1.125 cm.

Explanation:

The  refraction of light from the spherical surface is given by the formula:

n₂/v - n₁/u = [n₂ – n₁][1/R]

Where,

n₁ = Refractive index of air = 1 (known)

n₂ = Refractive index of glass = 1.5 (given)

u = Distance of the object = 4 cm - 1 cm = 3 cm

v = Distance of the image

R = Radius of the curvature = 4/2 = 2 cm

On substituting the values, we get,

1.5/v - 1/3 = (1.5 - 1)/2

1.5/v - 1/3 = 0.5 × 2

1.5/v - 1/3 = 1

1.5/v = 1 + 1/3

1.5/v = (3 + 1)/3

1.5/v = 4/3

v = 1.5 × 3/4 = 4.5/4

∴ v = Image distance = 1.125 cm