In a Young's double slit experiment, the slits are separated by 0.2 mm. If the slits separation is increased to 0.4 mm, the percentage change of the fringe width is:

Given: Initial slit separation is d1 = 0.2 mm and final slit separation is d2 = 0.4 mm. Find: The percentage change in fringe width. In Young's double slit experiment, fringe width is given by = Dd So, 1d. Therefore, the ratio of final and initial fringe widths is 2 1 = d1d2 = 0.20.4 = 12 Hence, 2 = 12 1 Now calculate percentage change: Percentage change = 2 - 1 1 100 Substituting 2 = 12 1, Percentage change = 12 1 - 1 1 100 = -50 The negative sign shows that the fringe width decreases. Therefore, the fringe width decreases by 50. The correct option is C.

In a Young's double slit experiment, the slits are separated by 0… — Solution | JEE Physics

MCQEasyJEE 2025Interference (Young's Experiment)

In a Young's double slit experiment, the slits are separated by $0.2 \, \text{mm}$ . If the slits separation is increased to $0.4 \, \text{mm}$ , the percentage change of the fringe width is:

A
$0\%$
B
$100\%$
C
$50\%$
D
$25\%$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Initial slit separation is $d_1 = 0.2 \, \text{mm}$ and final slit separation is $d_2 = 0.4 \, \text{mm}$ .

Find: The percentage change in fringe width.

In Young's double slit experiment, fringe width is given by

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

So, $\beta \propto \frac{1}{d}$ .

Therefore, the ratio of final and initial fringe widths is

\frac{\beta_2}{\beta_1} = \frac{d_1}{d_2} = \frac{0.2}{0.4} = \frac{1}{2}

Hence,

\beta_2 = \frac{1}{2}\beta_1

Now calculate percentage change:

\text{Percentage change} = \frac{\beta_2 - \beta_1}{\beta_1} \times 100

Substituting $\beta_2 = \frac{1}{2}\beta_1$ ,

\text{Percentage change} = \frac{\frac{1}{2}\beta_1 - \beta_1}{\beta_1} \times 100 = -50\%

The negative sign shows that the fringe width decreases. Therefore, the fringe width decreases by $50\%$ . The correct option is C.

Common mistakes

Using direct proportionality between fringe width and slit separation is incorrect because $\beta = \frac{\lambda D}{d}$ . As slit separation increases, fringe width decreases. Use inverse proportionality.
Reporting the answer as $-50\%$ instead of matching the option format can be misleading. The negative sign only indicates decrease, so the option should be the magnitude of decrease, $50\%$ .

Practice more Interference (Young's Experiment) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Interference (Young's Experiment)

Answer

Step-by-step solution

Common mistakes

Practice more Interference (Young's Experiment) questions

Related questions