Question:

In the figure, a ray of light is incident on a transparent liquid contained in a thin glass box at an angle of 45° with its one face. The emergent ray passes along the face AB. Find the refractive index of the liquid.
a ray of light is incident on a transparent liquid contained in a thin glass box

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The critical angle condition and refractive index relationship are key to understanding the behavior of light in different mediums, especially in designing optical instruments and understanding phenomena like total internal reflection.
Updated On: Feb 20, 2025
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Solution and Explanation

Given:  Angle of incidence on the glass box: 4545^\circ  The emergent ray passes along the face AB, indicating that the angle of refraction at the glass-liquid interface is 9090^\circ.  1. First Surface (Air-Glass Interface): sin45sinθ=μ \frac{\sin 45^\circ}{\sin \theta} = \mu Given sin45=12\sin 45^\circ = \frac{1}{\sqrt{2}}: 12=μsinθ \frac{1}{\sqrt{2}} = \mu \sin \theta 2. Second Surface (Glass-Liquid Interface): sin(90θ)sin90=1μ \frac{\sin(90^\circ - \theta)}{\sin 90^\circ} = \frac{1}{\mu} Since sin(90θ)=cosθ\sin(90^\circ - \theta) = \cos \theta and sin90=1\sin 90^\circ = 1: cosθ=1μ \cos \theta = \frac{1}{\mu} 3. Combining the Equations: 12sinθ=1cosθ \frac{1}{\sqrt{2} \sin \theta} = \frac{1}{\cos \theta} Simplifying: cosθ=2sinθ \cos \theta = \sqrt{2} \sin \theta tanθ=12 \tan \theta = \frac{1}{\sqrt{2}} 4. Finding sinθ\sin \theta: From the triangle GEF: sinθ=13 \sin \theta = \frac{1}{\sqrt{3}} 5. Calculating the Refractive Index (μ\mu): μ=12sinθ=12×13=32=32 \mu = \frac{1}{\sqrt{2} \sin \theta} = \frac{1}{\sqrt{2} \times \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}} Therefore, the refractive index of the liquid is: 32 \boxed{\sqrt{\frac{3}{2}}}

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