Question

question for JEE and neet aspirants only.. Full explanation is required.
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Answer 1

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Let a ray of light enter at A and the refracted beam is AB. Thus, the incident will be at an angle p.

For no refraction at the lateral face, p > C or, sin p > sin C

But p + r = 90° and p = (90° - r)

Therefore sin (90° - r) > sin C or cos r > sin C

The substitution for cos r can be found from Snell's law. Thus, according to the from Snell's law -

n = sin i/sin r

=> n sin r= sin i

=> cos r = √1−sin²r

=>√1−sin²i/n²

=>√1−sin²i/n² > sin C

=> 1−sin²i/n² > sin² C

Since the maximum value of sin i is 1.

now, 1 = √n²- 1/n = n

the n on the centre and n at the left get cancelled

=> n² - 1 = 1

Thus, the value of angle of incidence is n² = 2 or n = √2

so, the refractive index n must be greater than √2

thus, option (a)