Join WhatsApp Icon JEE WhatsApp Group
Instructions

In a Young's double slit experiment, each of the two slits $$A$$ and $$B$$, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 mm. The distance between the slits at time $$t$$ is given by $$d = (0.8 + 0.04 \sin \omega t)$$ mm, where $$\omega = 0.08$$ rad s$$^{-1}$$. The distance of the screen from the slits is 1 m and the wavelength of the light used to illuminate the slits is 6000 Å. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point $$O$$.

image
Question 31

The 8th bright fringe above the point $$O$$ oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer ($$\mu$$m), is ________.


Correct Answer: 601.50

The position of the $$m^{\text{th}}$$ bright fringe on the screen (measured from the central point $$O$$) in a Young’s double-slit experiment is given by the condition for constructive interference: $$y_m = \dfrac{m \, \lambda \, D}{d}$$, where

$$m$$ = fringe order, here $$m = 8$$,
$$\lambda$$ = wavelength of light, $$6000 \text{ Å} = 6000 \times 10^{-10}\,\text{m} = 6 \times 10^{-7}\,\text{m}$$,
$$D$$ = slit-screen distance, $$1\,\text{m}$$,
$$d$$ = instantaneous slit separation.

The slit separation oscillates as $$d = \bigl(0.8 + 0.04 \sin \omega t\bigr)\,\text{mm}$$ with $$\omega = 0.08\,\text{rad s}^{-1}$$. Hence

Maximum separation (when $$\sin \omega t = +1$$): $$d_{\max} = (0.8 + 0.04)\,\text{mm} = 0.84\,\text{mm} = 0.84 \times 10^{-3}\,\text{m}$$.
Minimum separation (when $$\sin \omega t = -1$$): $$d_{\min} = (0.8 - 0.04)\,\text{mm} = 0.76\,\text{mm} = 0.76 \times 10^{-3}\,\text{m}$$.

Because $$y_m$$ is inversely proportional to $$d$$, the fringe lies farthest from $$O$$ when $$d$$ is minimum and closest when $$d$$ is maximum.

Extreme positions of the 8th bright fringe:

$$y_{\max} = \dfrac{8 \times 6 \times 10^{-7} \times 1}{0.76 \times 10^{-3}} = \dfrac{4.8 \times 10^{-6}}{0.76 \times 10^{-3}} = \left( \dfrac{4.8}{0.76} \right) \times 10^{-3}\,\text{m} \approx 6.3158 \times 10^{-3}\,\text{m} = 6.3158\,\text{mm}$$.

$$y_{\min} = \dfrac{8 \times 6 \times 10^{-7} \times 1}{0.84 \times 10^{-3}} = \dfrac{4.8 \times 10^{-6}}{0.84 \times 10^{-3}} = \left( \dfrac{4.8}{0.84} \right) \times 10^{-3}\,\text{m} \approx 5.7143 \times 10^{-3}\,\text{m} = 5.7143\,\text{mm}$$.

Separation between the two extreme positions:

$$\Delta y = y_{\max} - y_{\min} = (6.3158 - 5.7143)\,\text{mm} = 0.6015\,\text{mm}$$.

Converting to micrometres: $$0.6015\,\text{mm} = 0.6015 \times 10^{3}\,\mu\text{m} = 601.5\,\mu\text{m}$$.

Hence, the required separation is 601.50 µm.

Get AI Help

Create a FREE account and get:

  • Free JEE Advanced Previous Papers PDF
  • Take JEE Advanced paper tests

50,000+ JEE Students Trusted Our Score Calculator

Predict your JEE Main percentile, rank & performance in seconds

Ask AI

Ask our AI anything

AI can make mistakes. Please verify important information.