The angles of incidence and refraction of a monochromatic ray of light of wavelength \lambda at an air-glass interface are i and r, respectively. A parallel beam of light with a small spread \delta\lambda in wavelength about a mean wavelength \lambda is refracted atthe same air-glass interface. The refractive index \mu of glass depends on the wavelength \lambda as \mu (\lambda) = {a+b}/ \lambda^2 where a and b are constants. Then the angular spread in the angle of refraction of the beam is

Correct Answer is : |2btanraλ3+bλδλ|

|{\sin i}/{\lambda^3{\cos r}}\delta\lambda|

|\frac{2b}{\lambda^3}\delta\lambda|

|\frac{2b \tan r}{a\lambda^3+ b\lambda}\delta\lambda|

|{2b({a + b}/\lambda^2) \sin i}/{\lambda^3}\delta\lambda|

KVPY SX SB Exam 20141 Question Paper

Question : 96 of 120

Marks: +1, -0

The angles of incidence and refraction of a monochromatic ray of light of wavelength

λ

at an air-glass interface are i and r, respectively. A parallel beam of light with a small spread

δ λ

in wavelength about a mean wavelength

λ

is refracted atthe same air-glass interface. The refractive index

µ

of glass depends on the wavelength

λ

µ (λ) = {a+b}/ λ^2

where a and b are constants. Then the angular spread in the angle of refraction of the beam is

Solution:

Snell law

\sin i = μ \sin r

\sin i =(a+b/λ^2)\sin r

Differentiating with respect to

λ

0 = \cos r dr (a+b/λ^2)

+\sinr(b/λ^3(-2))dλ

0 = \cos r dr ({aλ^2+b}/λ^2)

+\sin r({-2b}/λ^3)dλ

{dλ\ 2b\ \sin r}/λ=\cosr dr(aλ^2+ b)

d={2b \ dλ}/λ\ {\tan r}/(aλ^2+b)

δr={(2b \tan r)δr}/(aλ^3+bλ)

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