A light source, which emits two wavelength `lamda_(1)=400nm` and `lamda_(2)=600nm`, is used in a Young's double slit experiment. If recorded fringe width for `lamda_(1)` and `lamda_(2)` are `beta_(1)` and `beta_(2)` and the number of fringes for them within a distance y on one side of the central maximum are `m_(1)` and `m_(2)` respectively, then

Options

A. `beta_2 gt beta_1`
B. `m_1 gt m_2`
C. from the central maximum, `3^(rd)` maximum of `lambda_2` overlaps with `5^(th)` minimum of `lambda_1`
D. the angular separation of fringes for `lambda_1` is greater than `lambda_2`

Answer

A, B

Explanation

Fringe width is given by `beta = (lambda D)/d`
In the given case `lambda_2 > lambda_1 implies beta_2 > beta_1`, hence option (a) is correct.
More fringe width means less number of fringes can be accommodated in a given length. Hence `m_1 > m_2`. Hence option (b) is correct.
Let `m^(th)` of `lambda_1` matches with `n^(th)` of `lambda_2` then we can write the following equation:
`m (lambda_1D)/d = n (lambda_2D)/(d)`
`implies m/n = (lambda_2)/(lambda_1) = 600/400 = 3/2`
Third maximum of `lambda_1` matches with second maximum of `lambda_2`. Hence option (c ) is wrong.
Angular fringe width can be written as `theta = beta/D = lambda/d`.
Therefore, angular separation is less for `lambda_1` than `lambda_2`. hence option (d) is wrong.