Question

The intensity at the maximum in a Young's double slit experiment is $I_0$. 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 at a distance $D=10d$ ?

• A. $I_0$
• B. $\frac{I_0}{4}$
• C. $\frac{3}{4}{I}_0$
• D. $\frac{I_0}{2}$

Answer 1

The correct option is (D) I02\cfrac { { I }_{ 0 } }{ 2 } 2I0​​

Step 1, Given data

Distance between the two slits (d) = $5\lambda$

Distance of screen = 10d

Intensity at the maximum = $I_0$

We have to find the resultant intensity at $y=\frac{d}{2}$

Step 2, Finding path difference

We know,

Path difference, $\Delta x=dtan\theta =\frac{dy}{D}$

Now putting all the values

$\Delta x=\frac{d^2}{2D}=\frac{d^2}{2\times 10d}=\frac{d}{20}$

Putting value of d

$\Delta x=\frac{5\lambda }{20}=\frac{\lambda }{4}$

Step 3, Finding the resultant intensity

We know,

$\theta =\frac{2\pi }{\lambda }\times \Delta x=\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$

From the equation resultant intensity

$I_y=I_0{cos}^2\left(\frac{\theta }{2}\right)$

Putting all the values

$I_y=I_0{cos}^2\left(\frac{\pi }{4}\right)=\frac{I_0}{2}$

Hence, resultant intensity is $\frac{I_0}{2}$

The correct option is (D)

Path difference, Δx=dtan⁡θ=dyD=d22D=d22×10d=d20=λ4$$\begin{array}{cc}\text{Path difference,}\Delta x=d\tan \theta =\frac{dy}{D}=\frac{d^2}{2D}& =\frac{d^2}{2\times 10d}\\ & =\frac{d}{20}\\ & =\frac{\lambda }{4}\end{array}$$