Question

In Young's double slit experiment with two different set-ups, the fringe widths are equal. If the ratio of slit separation is $2$ and the ratio of wavelengths is $\dfrac{1}{2}$, find the ratio of screen distances in both set-ups:

• A. $\dfrac{D_1}{D_2}=3$
• B. $\dfrac{D_1}{D_2}=2$
• C. $\dfrac{D_1}{D_2}=8$
• D. $\dfrac{D_1}{D_2}=4$

Hint:

If fringe width remains unchanged in YDSE problems, directly equate $\dfrac{\lambda D}{d}$ for both cases to avoid unnecessary calculations.

Answer 1

The Correct Option is D

In Young's double-slit experiment, the fringe width \(\beta\) is given by the formula:

\(\beta = \frac{\lambda D}{d}\)

where:

• \(\lambda\) is the wavelength of the light used.
• \(D\) is the distance between the slits and the screen.
• \(d\) is the separation between the slits.

Given that the fringe widths are equal for the two setups, we can say:

\(\beta_1 = \beta_2\)

Thus,

\(\frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2}\)

We are provided with:

• The ratio of slit separation \(\frac{d_1}{d_2} = 2\).
• The ratio of wavelengths \(\frac{\lambda_1}{\lambda_2} = \frac{1}{2}\).

To find the ratio of the screen distances \(\frac{D_1}{D_2}\), substitute the known ratios into the equation. So, we have:

\(\frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2}\)

Substitute \(\frac{d_1}{d_2} = 2\) and \(\frac{\lambda_1}{\lambda_2} = \frac{1}{2}\):

\(\frac{\frac{1}{2} \lambda_2 D_1}{2 d_2} = \frac{\lambda_2 D_2}{d_2}\)

Cancel \(\lambda_2\) and \(d_2\), and simplify:

\(\frac{D_1}{4} = D_2\)

This implies:

\(D_1 = 4 D_2\)

Therefore, the ratio of the screen distances is:

\(\frac{D_1}{D_2} = 4\)

Thus, the correct answer is:

\(\frac{D_1}{D_2}=4\)