question_answer

In the Youngs double slit experiment, the intensities at two points \[{{P}_{1}}\] and \[{{P}_{2}}\] on the screen are respectively \[{{I}_{1}}\] and \[{{I}_{2}}\]. If \[{{P}_{1}}\] is located at the centre of a bright fringe and \[{{P}_{2}}\] is located at a distance equal to a quarter of fringe width from \[{{P}_{1}}\], then \[\frac{{{I}_{1}}}{{{I}_{2}}}\] is

A)  2                 

B)  \[\frac{1}{2}\]

C)  4                

D)  16

Correct Answer: D

Solution :

Fringe width    \[\beta =\frac{\lambda D}{d}\] Let the amplitude of that place where constructive inference takes place is a.                   The position of fringe at \[{{p}_{2}}\] is \[x=\frac{n\lambda D}{d}\] Given,          \[\beta =\left( \frac{\beta }{4} \right)\] \[\therefore \] \[\frac{\lambda D}{4d}=\frac{n\lambda D}{d}\] or \[n=\frac{1}{4}\] \[\therefore \] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{a}^{2}}}{{{\left( \frac{a}{4} \right)}^{2}}}\] or \[{{I}_{1}}:{{I}_{2}}=16:1\]