Question

In young’s double-slit experiment the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$, ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\dfrac{\lambda}{4}$​ will be
A). $K$
B). $\dfrac{K}{4}$
C). $\dfrac{K}{2}$
D). $Zero$

Answer 1

The intensity of lightwave at a point and the phase difference is related to each other. Here path difference is given. We have the formula for phase difference in terms of path difference. Using the formula first, find the phase difference and then the intensity of light.
Formula used:
${{I}^{‘}}=4{{I}_{0}}{{\cos }^{2}}\left( \dfrac{\phi }{2} \right)$

Complete step-by-step solution:
We have, net intensity of light at any point on screen,
${{I}^{‘}}=4{{I}_{0}}{{\cos }^{2}}\left( \dfrac{\phi }{2} \right)$------- (1)
Where,
${{I}_{0}}$ is the intensity of wave
$\phi$ is the phase difference between two waves.
Given that, $I’=K$ ---------(2)
We have,
$\phi =\dfrac{2\pi }{\lambda }x$ -------- (3)
Where,
$\phi$ is the phase difference
$x$ is the path difference.
$\lambda$ is the wavelength.
Substitute $x=\lambda$ in equation 3. Then,
$\phi =\dfrac{2\pi }{\lambda }\times \lambda =2\pi$ ----- (4)
Substitute 2, 4 in equation 1, we get,
$K=4{{I}_{0}}{{\cos }^{2}}\left( \dfrac{2\pi }{2} \right)=4{{I}_{0}}{{\cos}^{2}}\left( \pi \right)=4{{I}_{0}}$ ----------- (5)
For the second case, where,
$x=\dfrac{\lambda }{4}$
Substitute $x=\dfrac{\lambda }{4}$ in equation 3. Then,
$\phi =\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{4}=\dfrac{\pi }{2}$ ----- (6)
Substitute 2, 6 in equation 1, then,
${{\text{K}}^{‘}}\text{=4}{{\text{I}}_{0}}\text{co}{{\text{s}}^{\text{2}}}\left(\dfrac{\dfrac{\text{}\\!\\!\pi\\!\\!\text{ }}{\text{2}}}{\text{2}} \right)=4{{I}_{0}}{{\cos }^{2}}\left( \dfrac{\pi }{4} \right)=2{{I}_{0}}$
$\Rightarrow {{K}^{‘}}=2{{I}_{0}}$
Then, from equation 5,
$K=4{{I}_{0}}$
Therefore,
${{K}^{‘}}=\dfrac{K}{2}$
Hence the answer is option C.

Note: Intensity of lightwave at a point is proportional to the square root of its amplitude. This means that, if we want to emulate the effect of a sound being twice as far away, then we would need to multiply the amplitude by one-half. From the inverse square law, we can understand that the amplitude of a light wave is inversely proportional to distance.