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CBSE Class 12 Physics 2019 Delhi Set 1 Paper

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Question : 4 of 27
Marks: +1, -0
What is the speed of light in a denser medium of polarising angle 30∘30^{\circ} ?
Solution:  
dμr=tan⁡30∘=13d \mu_r = \tan 30^{\circ} = \frac{1}{\sqrt{3}}
(where dμrd \mu_r, is the refractive index of rarer medium w.r.t. denser medium)
∴μd=3\therefore \mu_d = \sqrt{3}
v=cμ=3×1083=3×108 m/sv = \frac{c}{\mu} = \frac{3 \times 10^8}{\sqrt{3}} = \sqrt{3} \times 10^8 \, \text{m} / \text{s}
[Note : Also accept if a student solves it as follows]
μ=tan⁡ip\mu = \tan i_p
μ=tan⁡30∘=13\mu = \tan 30^{\circ} = \frac{1}{\sqrt{3}}
∴v=3×10813=33×108 m/s\therefore v = \frac{3 \times 10^8}{\frac{1}{\sqrt{3}}} = 3\sqrt{3} \times 10^8 \, \text{m} / \text{s}
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