Question

The intensity at the maximum in a Young’s double slit experiment is \[{I_o}\].Distance between two slits is \[d = 5\lambda \], where \[\lambda \]is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D=10d?

Answer 1

Hint: You need to calculate path difference and then phase difference using path difference to calculate intensity in front of one of the slits on the screen or else you can simply put formulas but here measurements given are not very ideal so solving with basics will be more accurate.

Complete step by step solution:

Let intensity of light due to each slit is \[I\]. Then at the centre, maximum intensity will occur and its value will be \[4I\]. Because at centre phase difference is zero and if we use the formula

${I_{net}} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \Phi \\

\Rightarrow {I_ \circ } = I + I + 2\sqrt {I \times I} \cos 0 = 4I $

Therefore \[I = \dfrac{{{I_o}}}{4}\]

Now we will find path difference at point in front of one of the slits and distance travelled by light from one of the slit is 

\[{s_1} = D = 10d = 50\lambda \]

And distance travelled by light from other slit is

${s_2} = \sqrt {{d^2} + {D^2}} \\

\Rightarrow {s_2} = \sqrt {{d^2} + 100{d^2}} \\

\Rightarrow {s_2} = \sqrt {101} d \\

\Rightarrow {s_2} = 10.05d \\

\Rightarrow {s_2} = 50.25\,\lambda$

So path difference is ${s_2} - {s_1} = 50.25\,\lambda - 50\,\lambda = 0.25\,\lambda = \dfrac{\lambda}{4}$

Now we know that phase difference is $\dfrac{2\pi}{\lambda} \times\,\text{path difference}$

$\text{Phase difference} = \dfrac{2\pi}{\lambda} \times\,\text{path difference}$

$\Rightarrow \text{Phase difference} = \dfrac{2\pi}{\lambda} \times\,\dfrac{\lambda}{4}$

$\Rightarrow \text{Phase difference} = \dfrac{\pi}{2}$

Value of cos of $\dfrac{\pi}{2}$ will be zero.Therefore new intensity will be

\[{I_{net}} = I + I + 2\sqrt {I \times I}\, \cos {\dfrac{\pi}{2}}\\

\Rightarrow {I_{net}} = I + I + 2\sqrt {I \times I} \times 0 \\

\Rightarrow {I_{net}} = 2I \\

\Rightarrow {I_{net}} = 2\dfrac{{{I_o}}}{4} \\

\therefore {I_{net}} = \dfrac{{{I_o}}}{2}\]

Hence new intensity in front of one of the slits will be \[\dfrac{{{I_o}}}{2}\].

Note:
Solve using basics and we got some finite non zero value of intensity but if we had solved it with formulas we would have got intensity as zero because according to formulas there will be minima in front of any of the slits as distance from centre is odd multiple of half of wavelength. This happens because measurements here are not ideal and you should always look for them.