Question

In Young's double slit experiment with monochromatic light of wavelength 600 nm , the distance between the slits is \(10^{-3} \mathrm{~m}\). For changing the fringe width by \(3 \times 10^{-5} \mathrm{~m}\)

a. the screen is moved away from the slit by 5 cm .

b. the screen is moved 5 cm towards the slits.

c. the screen is moved 3 cm towards the slits.

d. the screen is moved away from the slits by 3 cm .

• A. both (c) and (d)
• B. both (a) and (b)
• C. only (a)
• D. only (c)

Hint:

Fringe width is proportional to the screen distance, so a change of 3e-5 m in fringe width needs a 5 cm shift either way.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Fringe width \(\beta=\dfrac{\lambda D}{d}\), so \(\Delta\beta=\dfrac{\lambda\,\Delta D}{d}\).

Step 2: Find the shift
\(\Delta D=\dfrac{\Delta\beta\,d}{\lambda}=\dfrac{3\times10^{-5}\times10^{-3}}{6\times10^{-7}}=0.05\text{ m}=5\text{ cm}\).

Step 3: Direction
Moving the screen away increases the fringe width by \(3\times10^{-5}\) m; moving it 5 cm closer decreases it by the same amount. Both change the width by that magnitude.

Step 4: Compare the options
So statements (a) and (b) are correct.

Final Answer:
Both (a) and (b) change the fringe width by 3 times 10 to the power minus 5 metres. \[ \boxed{\text{(B) }\text{both (a) and (b)}} \]