Question

A solid glass sphere of refractive index $ n = \sqrt{3} $ and radius $ R $ contains a spherical air cavity of radius $ \dfrac{R}{2} $, as shown in the figure. A very thin glass layer is present at the point $ O $ so that the air cavity (refractive index $ n = 1 $) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source $ S $ emits a light ray from a point inside the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point $ O $ and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is $ \theta $. The value of $ \sin \theta $ is ____ 

Hint:

For curved geometries, draw auxiliary triangles from center and use trigonometric relations like \( \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} \). This often helps when direct formulae like Brewster's law are not sufficient.

Answer 1

Correct Answer: 0.75 - 0.86

To find the value of \(\sin \theta\), we need to analyze the optics problem involving a solid glass sphere with a spherical air cavity, as shown in the diagram. Step 1: Understand the setup

• The glass sphere has a refractive index \( n = \sqrt{3} \) and radius \( R \).
• The air cavity inside the sphere has a radius \( \frac{R}{2} \) and is centered at point \( O \), which is also the point where a very thin glass layer separates the air cavity from the glass.
• The light source \( S \) emits unpolarized, unidirectional, and monochromatic light from inside the air cavity toward the periphery of the glass sphere.
• The light is reflected at point \( O \) (the interface between the air cavity and the glass sphere), and the angle of incidence at this inner surface is \( \theta \).
• The reflected light is fully polarized, which implies that the angle of incidence \( \theta \) must be the Brewster angle (also called the polarizing angle).

Step 2: Identify the Brewster angle condition The Brewster angle occurs when the reflected light is fully polarized, which happens when the angle of incidence satisfies the condition: \[ \tan \theta = \frac{n_{\text{glass}}}{n_{\text{air}}} \] Here:

• \( n_{\text{glass}} = \sqrt{3} \) (refractive index of the glass),
• \( n_{\text{air}} = 1 \) (refractive index of the air inside the cavity).

So, the Brewster angle condition becomes: \[ \tan \theta = \frac{\sqrt{3}}{1} = \sqrt{3} \] 
Step 3: Calculate \( \sin \theta \) From the Brewster angle condition, we have: \[ \tan \theta = \sqrt{3} \] We know from trigonometry that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), and using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can find \(\sin \theta\). Since \(\tan \theta = \sqrt{3}\), we can think of a right triangle where:

• The opposite side (related to \(\sin \theta\)) is \(\sqrt{3}\),
• The adjacent side (related to \(\cos \theta\)) is \(1\),
• The hypotenuse is \(\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2\).

Thus: \[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} \] Alternatively, we can use the identity: \[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\sin^2 \theta}{1 - \sin^2 \theta} \] Given \(\tan \theta = \sqrt{3}\), we have: \[ (\sqrt{3})^2 = \frac{\sin^2 \theta}{1 - \sin^2 \theta} \] \[ 3 = \frac{\sin^2 \theta}{1 - \sin^2 \theta} \] Let \( x = \sin^2 \theta \). Then: \[ 3 (1 - x) = x \] \[ 3 - 3x = x \] \[ 3 = 4x \] \[ x = \frac{3}{4} \] \[ \sin^2 \theta = \frac{3}{4} \] \[ \sin \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] Since \(\theta\) is the angle of incidence (between \(0^\circ\) and \(90^\circ\)), \(\sin \theta\) is positive. 
Step 4: Final answer The value of \(\sin \theta\) is: \[ \sin \theta = \frac{\sqrt{3}}{2} \] 
Final Answer: \( \boxed{\sin \theta = 0.86} \)