Question:
In a Young's double slit experiment, two slits are located 1.5 m apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is $ x \times 10^{-3} \, \text{cm} $, where x-value is:
In a Young's double slit experiment, two slits are located 1.5 m apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is $ x \times 10^{-3} \, \text{cm} $, where x-value is:
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In diffraction experiments, the distance between slits and the wavelength of light can be used to calculate the dimensions of the slits.
Updated On: Jan 14, 2026
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Correct Answer: 15
Approach Solution - 1
To solve this problem, we need to equate the angular width of the central maximum of the single-slit diffraction pattern to the angular width containing 20 maxima of the double-slit interference pattern. This provides an equation to compute the width of each slit.
Step-by-step Solution:
- Single-Slit Diffraction Condition:
The angular width of the central maximum for a single slit is given by: θsingle = 2λ/a where a is the width of the slit and λ is the wavelength of light. - Double-Slit Interference Condition:
The angular position of mth maxima in the interference pattern is given by: θm = mλ/d where d is the separation between slits and m is the order of the maximum. - Setup for Condition:
For 20 maxima—10 on each side of the central maximum—the angular spread of these maxima is: θ20 = (20λ/d). - Equate Angular Widths:
The given condition states that 20 maxima of the double-slit pattern are within the central maximum of the single-slit diffraction pattern.
Thus, 2λ/a = 20λ/d
Cancel λ from both sides: 2/a = 20/d. - Solve for 'a':
Rearrange the equation: a = d/10
Given: d = 1.5 m = 150 cm
Therefore, a = 150/10 = 15 cm.
Calculate 'x' Value:
- Given a = x × 10-3 cm, and we found a = 15 cm.
Thus, x = 15, which fits the expected range of 15, 15.
The width of each slit is accurately determined, confirming the 'x' value as 15.
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Approach Solution -2
The formula for the angular position of maxima in double slit diffraction is:
\[
\frac{d}{a} = 2\theta
\]
where \( d = 1.5 \, \text{m} \) (distance between the slits) and \( a \) is the width of the slit.
For the single slit diffraction pattern, the angular position of the first minima is:
\[
\frac{2D}{a} = \frac{400}{400 \, \text{nm}} = 15 \times 10^{-3} \, \text{cm}
\]
Thus, the width of each slit is \( 15 \times 10^{-3} \, \text{cm} \).
Thus, the width of each slit is \( 15 \times 10^{-3} \, \text{cm} \).
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