A point source of light is placed 4 m below the surface of water of refractive index is 5/3. The minimum diameter of a disc which should be placed over the source on the surface of water to cut-off all light coming out of water is:

Question

A point source of light is placed 4 m below the surface of water of refractive index is 5/3. The minimum diameter of a disc which should be placed over the source on the surface of water to cut-off all light coming out of water is: (A) 6m (B) 3 m (C) 4m (D) 2 m

Tardigrade Admin · Accepted Answer

Given, sin i=(1/μ )=(3/5), so tan i=(3/4)=(r/4) it gives r = 3 m so, the diameter = 6 m

Q. A point source of light is placed 4 m below the surface of water of refractive index is 5/3. The minimum diameter of a disc which should be placed over the source on the surface of water to cut-off all light coming out of water is:

6m

3 m

4m

2 m

Given, $sin i = \frac{1}{μ} = \frac{3}{5},$ so $tan i = \frac{3}{4} = \frac{r}{4}$ it gives r = 3 m so, the diameter = 6 m

Q. A point source of light is placed 4 m below the surface of water of refractive index is 5/3. The minimum diameter of a disc which should be placed over the source on the surface of water to cut-off all light coming out of water is:

6m

3 m

4m

2 m

Given, sini=μ1​=53​, so tani=43​=4r​ it gives r = 3 m so, the diameter = 6 m

Given, $sin i = \frac{1}{μ} = \frac{3}{5},$ so $tan i = \frac{3}{4} = \frac{r}{4}$ it gives r = 3 m so, the diameter = 6 m