Two parallel beams of light P and Q (separation d) containing radiation of wavelengths `4000A` and `5000A` (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in fig. The refractive index of the prism as a function of wavelength is given by the relation. `mu(lamda)=1.20+(b)/(lamda^(2))` Where `lamda` is in `A` and b is positive constant. The value of b is such that the condition wave length and is not satisfied for the other.
(a) Find the value of b.
(b) find the deviation of the beams transmitted through the face AC. (c) A convergent lens is used to bring these transmitted beams into focus. If the intensities of transmission form the face AC, are 41 and I respectively, find the resultant intensity at the focus. `

Answer

A, B, C, D

Explanation

(a) `lamda_(1)=4000A` and `lamda_(2)=5000A`
For total internal reflection to take place, `theta` should be greater than C.For smaller values of C, the values of `mu` should be high or in other words the value of `lamda` should be small.
Therefore, total internal reflection will be given by `lamda_(1)=4000A`
Here,`sintheta=0.8` (given) `implies` `theta=53.1^circ` `:'` `mu=(1)/(sintheta)=(1)/(0.8)=1.25`
`:'``mu=1.2+(b)/((400xx10^(-10))^(2))=1.25` `implies` `b=0.8xx10^(-14)m^2`
(b) Applying Snell's law at N for wavelength `lamda_(2)`
`mu=(sinr)/(sini)`where`mu=1.5+(0.8xx10^(-14))/((5000xx10^(-10))^(2))=1.232` `implies``1.232=(sinr)/(0.8)impliesr=80.3^circ`
From the figure it is clear that the deviation, `delta=r-i=80.3^circ-53.1^circ=27.2^circ`
(c) The intensities of transmitted beams are 4I and I respectively
Path diff=`mu(MB)-AN`=`(sinr)/(sini)(ABsini)-ABsintheta`
`=0`
Since both the radiation are mutually coherent and while coming to focus these travel equal paths, therefore, these two beams will arrive in phase at focus. `:'` Resultant Intensity
`I_(max)=(sqrtI_(1)+sqrtI_(2))^(2)=(sqrt4I+sqrtI)^(2)=(3sqrtI)^(2)=9I`.