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

Concept: In Young’s double slit experiment, the fringe width $\beta$ is given by: \[ \beta = \frac{\lambda D}{d} \] where $\lambda$ = wavelength of light, $D$ = distance of screen from slits, $d$ = slit separation.
Step 1: Since fringe widths in both set-ups are equal: \[ \beta_1 = \beta_2 \] \[ \frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2} \]
Step 2: Rearranging: \[ \frac{D_1}{D_2} = \frac{d_1}{d_2} \cdot \frac{\lambda_2}{\lambda_1} \]
Step 3: Given: \[ \frac{d_1}{d_2} = 2, \quad \frac{\lambda_1}{\lambda_2} = \frac{1}{2} \Rightarrow \frac{\lambda_2}{\lambda_1} = 2 \] Step 4: Substitute the values: \[ \frac{D_1}{D_2} = 2 \times 2 = 4 \]
Step 5: Hence, \[ \boxed{\frac{D_1}{D_2}=4} \]