The angles of incidence and refraction of a monochromatic ray of light of wavelength `lamda` at an air-glass interface are i and r, respectively. A parallel beam of light with a small spread `delta lamda` in wavelength about a mean wavelength `lamda` is refracted at the same air-glass interface. The refractive index `mu` of glass depends on the wavelength `lamda` as `mu(lamda) = a + b//lamda2` where a and b are constants. Then the angular spread in the angle of refraction of the beam is

Options

A. `|(sin i )/(lambda^(3) cosr)delta lambda|`
B. `|(2b)/(lambda^(3))delta lambda|`
C. `|(2b tanr)/(alambda^(3)+ b lambda)delta lambda|`
D. `|(2b(a+b// lambda^(2) sini ))/(lambda^(3))delta lambda |`

Answer

C

Explanation

Snell.s law
`sin i = mu `sin r
`rArr " sin i = (a + (b)/(lambda^(2)) ) ` sin r
Differentiating with respect to `lambda`
O = cos r dr `(a + (b)/(lambda^(2))) + sin r( (b)/(lambda^(3)) (-2) ) d lambda`
`rArr " " (d lambda 2 b sin r)/(lambda) = cos r dr( a lambda^(2) + b)`
`rArr dr = (2 b d lambda)/(lambda) (tan r )/((a lambda^(2) + b))`
`delta r = ((2b tan r) delta lambda)/((a lambda^(3) + b lambda))`.