Question

In double slit experiment, the two slits are \(1\) mm apart and the screen is placed \(1\) m away. A monochromatic light of wavelength \(500\) nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

• A. \(0.2\) mm
• B. \(0.1\) mm
• C. \(0.5\) mm
• D. \(0.02\) mm

Hint:

Equate the central single-slit width 2 lambda D / a to ten double-slit fringe widths lambda D / d.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The central maximum of a single slit of width \(a\) has linear width \(\dfrac{2\lambda D}{a}\).
The double slit fringe width is \(\beta=\dfrac{\lambda D}{d}\).

Step 2: Condition for ten maxima
Ten double-slit maxima fit inside the central maximum, so
\(\dfrac{2\lambda D}{a}=10\,\dfrac{\lambda D}{d}\)

Step 3: Calculation
\(a=\dfrac{2d}{10}=\dfrac{d}{5}\)
With \(d=1\) mm, \(a=0.2\) mm.

Step 4: Compare the options
0.2 mm matches option (A).

Final Answer:
Each slit should be 0.2 mm wide. \[ \boxed{\text{(A) }0.2\text{ mm}} \]