Question

16. A ray of light travels from air into a glass block as shown. It makes an angle of 30° degrees with the surface of the block. If the refractive index of the glass is 1.5, what is the angle of refraction? Air (A) (B) (C) (D) 1.30° 19.47° 35.26° 48.590 30° Glass block​

Answer 1

Explanation:

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A ray of light travels from air into a perspex block of refractive index 1.5 at an angle of incidence of 45o

Calculate the angle of the refraction of the ray at air-to-perspex boundary and at perspex-to-air boundary.

Solution:

Given:

Refractive index of block μ

=1.5

Angle of incidence i=45°

For refraction from air to block:

We apply Snell's law,

μ aair sin i=μ block sin r

1×sin 45° = 1.5×sin r

sin r= sin 45°/1.5 . . . (i)

On simplifying,

sin r= (1/✓2)/1.5

sin r=(✓2/3)

Answer : r=28.12° is the angle of the refraction of the ray at air-to-perspex boundary.

Now when the ray will move from block to air, then r is incidence angle.

Applying Snell's law again,

μ bblock sinr=μ aair sinr 1

Using equation (i) to substitute the value of sin r,

1×sin r1 X = μ bbloc X sin r

sinr 1 =1.5× sin 45°/1.5 ⇒sin rr =sin 45°

Answer : rr = 45° is the angle of the refraction of the ray at perspex-to-air boundary.

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