Question

In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is $5\mathrm{mm}$. The screen on which the diffraction pattern is displayed is at a distance of $80~\mathrm{cm}$ from the slit. The wavelength is $6000\mathrm{\AA}$. The slit width in (mm) is about

• A. $0.576$
• B. $0.348$
• C. $0.192$
• D. $0.096$

Hint:

Slit width for single-slit diffraction: $w = \frac{D\lambda}{y_1}$, where $y_1$ is the distance from the centre to the first minimum.

Answer 1

The Correct Option is C

The given problem involves a single-slit diffraction pattern. To find the width of the slit, we can use the formula for the angular position of the minima in a diffraction pattern:

\(a \sin \theta = m\lambda\)

where:

• \(a\) is the slit width,
• \(\theta\) is the angle of the minima from the central maximum,
• \(m\) is the order of the minima (for the first minima, \(m = \pm 1\)),
• \(\lambda\) is the wavelength of light.

The problem gives the total distance between the first minima on either side of the central maximum, which is 5 mm. We know the relationship:

\(\Delta y = \frac{2\lambda L}{a}\)

where:

• \(\Delta y\) is the distance between the first minimum on the left and right,
• \(L\) is the distance from the slit to the screen.

Given values:

• \(\Delta y = 5 \, \mathrm{mm} = 5 \times 10^{-3} \, \mathrm{m}\)
• \(\lambda = 6000 \, \mathrm{\AA} = 6000 \times 10^{-10} \, \mathrm{m}\)
• \(L = 80 \, \mathrm{cm} = 0.8 \, \mathrm{m}\)

Substituting these into the equation, we have:

\(5 \times 10^{-3} = \frac{2 \times 6000 \times 10^{-10} \times 0.8}{a}\)

Solving for \(a\):

\(a = \frac{2 \times 6000 \times 10^{-10} \times 0.8}{5 \times 10^{-3}} = \frac{9600 \times 10^{-10}}{5 \times 10^{-3}}\)

\(a = \frac{9600 \times 10^{-10}}{5 \times 10^{-3}} = 1.92 \times 10^{-4} \, \mathrm{m} = 0.192 \, \mathrm{mm}\)

Hence, the slit width is about 0.192 mm, which matches the provided correct option.