Question

In Young’s double slit experiment, the intensity of light at a point on the screen where the path difference λ is K, (λ being the wavelength of the 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

Hint: Young’s double slit experiment was demonstrated to observe the phenomenon of interference pattern. We can solve this question by using the intensity formula and applying the phase difference equation.

Formula used: Intensity, $I=4{ I }_{ 0 }{ cos }^{ 2 }\dfrac { \phi }{ 2 }$
Phase difference, $\Phi =\dfrac { 2\pi }{ \lambda }$× path difference

Complete step by step solution:
Intensity is the power of light passing through a unit area or the amount of energy arriving per unit time. It is denoted as ‘I’ and is given by the formula,

$I=4{ I }_{ 0 }{ cos }^{ 2 }\dfrac { \phi }{ 2 }$

Where, I₀ is the intensity of the other wave and
     Φ is the phase difference between two waves
The phase difference describes the difference in the lag of two waves. It is denoted as ‘ϕ’ and given by the formula,

$\Phi =\dfrac { 2\pi }{ \lambda }$× path difference
The given path difference is λ, then phase difference can be written as
$\Phi =\dfrac { 2\pi }{ \lambda } \times \lambda$

Both λ gets cancelled and we get, ϕ = 2π

Now, let us substitute the value of ϕ in the intensity formula,

$I=4{ I }_{ 0 }{ cos }^{ 2 }\left( \dfrac { 2\pi }{ 2 } \right)$
$I=4{ I }_{ 0 }{ cos }^{ 2 }\left( \pi \right)$
$I=4{{I}_{0}}=K$
(K as assumed in the question)
Now, consider the path difference is λ/4, then the phase difference becomes
$\phi =\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{4}=\dfrac{\pi }{2}$
Now substituting the above value of ϕ in the intensity formula. We get,

$I=4{ I }_{ 0 }{ cos }^{ 2 }\left( \dfrac { \pi }{ 4 } \right)$
$I=4{{I}_{o}}=\dfrac{K}{2}$ (since I₀ = K)
Therefore, the correct answer for the given question is option (C).

Note: Make clear note of the given λ and ϕ values. Path difference of $\dfrac{\lambda }{4}$ corresponds to the phase difference of $\dfrac{\pi }{2}$. In the same way, the path difference corresponding to phase difference of π is $\dfrac{\lambda }{2}$.