Question

A glowing bulb is placed at a depth root 7 meter ina pond of water( refractive index 4/3) what is the minimum diameter of the opaque circular disc placed on the surface of the water so that the bulb is not visble

Answer 1

Answer:

15.96 m

Explanation:

Actual depth of the bulb in water (d1) = 7 m

Refractive index of water (μ) = 1.33

from the figure

Where,

Angle of incidence  =  Angle of refraction = 90°

Sincethe bulb is a point source, the emergent light can be considered as a circle of radius

R = MP/2 = MO = OP

μ = sin r/ sini

1.33 = sin 90°/sini

i = sin-1 (1/1.33) = 48.75°

From the figure

tan i OP/ON = R/d1

R = tan 48.75° x 7 = 7.98 m

Diameter= 2*R= 15.96 m