© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Determine the angle of refraction at the first surface
When the light ray enters the sphere from air, apply Snell’s law:
n_1 \sin i = n_2 \sin r
Here,
n_1 = 1 (refractive index of air)
i = 60^\circ (angle of incidence)
n_2 = \sqrt{3} (refractive index of the sphere)
Substituting:
1 \cdot \sin(60^\circ) = \sqrt{3} \cdot \sin(r)
Since \sin(60^\circ) = \frac{\sqrt{3}}{2} ,
\frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin(r) \quad \Rightarrow \quad \sin(r) = \frac{1}{2} \quad \Rightarrow \quad r = 30^\circ
Step 2: Determine the angle of incidence at the second (farther) surface
By the geometry of refraction through a sphere (and symmetry considerations), the angle at which the ray inside the sphere strikes the farther surface is the same as its refraction angle at the first surface. Thus, the angle of incidence at the second boundary is 30^\circ .
Step 3: Find the angle of refraction at the second surface
Again, apply Snell’s law at the second surface (from glass to air):
n_{\text{glass}} \sin(i_2) = n_{\text{air}} \sin(r_2)
Here,
n_{\text{glass}} = \sqrt{3}
i_2 = 30^\circ (from Step 2)
n_{\text{air}} = 1
r_2 is the refraction angle into air
Substituting:
\sqrt{3} \cdot \sin(30^\circ) = 1 \cdot \sin(r_2)
Since \sin(30^\circ) = \frac{1}{2} ,
\sqrt{3} \times \frac{1}{2} = \sin(r_2) \quad \Rightarrow \quad \sin(r_2) = \frac{\sqrt{3}}{2} \quad \Rightarrow \quad r_2 = 60^\circ
Step 4: Determine the angle of reflection at the second surface
The reflection angle at any boundary equals the incidence angle (law of reflection). Hence, the reflected ray inside the sphere makes an angle of 30^\circ with the normal at the second surface.
Step 5: Calculate the angle between the reflected and refracted rays
The reflected ray is 30^\circ from the normal (inside the sphere), and the refracted ray is 60^\circ from the normal (outside the sphere). Therefore, the angle between them is
30^\circ + 60^\circ = 90^\circ.
Hence, the angle between the reflected and refracted rays at the farther surface is 90^\circ .
