Question

A light source of 5000Å wavelength produces a single slit diffraction. The first minima in diffraction pattern is seen, at a distance of 5mm from central maxima. The distance between screen and slit is 2 metre. The width of slit in mm will be :

• A. 0.1
• B. 0.2
• C. 0.4
• D. 2

Answer 1

Answer: (B)

0.2

Answer 2

The problem involves single-slit diffraction. The position of the minima in a single-slit diffraction pattern is given by the formula:

$x_n = \frac{n \lambda D}{a}$

where:

$x_n$ is the distance of the nth minima from the central maxima

$n$ is the order of the minima (n = 1 for the first minima, n = 2 for the second minima, etc.)

$\lambda$ is the wavelength of light

$D$ is the distance between the slit and the screen

$a$ is the width of the slit

Given values:

Wavelength of light, $\lambda = 5000 \text{ Å} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m}$

Distance of the first minima from the central maxima, $x_1 = 5 \text{ mm} = 5 \times 10^{-3} \text{ m}$

Distance between the screen and the slit, $D = 2 \text{ metre} = 2 \text{ m}$

For the first minima, $n = 1$.

We need to find the width of the slit, $a$. Rearranging the formula for $n=1$:

$x_1 = \frac{1 \cdot \lambda D}{a}$

$a = \frac{\lambda D}{x_1}$

Substitute the given values into the formula:

$a = \frac{(5 \times 10^{-7} \text{ m}) \times (2 \text{ m})}{(5 \times 10^{-3} \text{ m})}$

$a = \frac{10 \times 10^{-7}}{5 \times 10^{-3}}$

$a = 2 \times 10^{-7} \times 10^{3}$

$a = 2 \times 10^{-4} \text{ m}$

The question asks for the width of the slit in millimeters (mm). Convert meters to millimeters:

$1 \text{ m} = 1000 \text{ mm}$

$a = 2 \times 10^{-4} \text{ m} \times \frac{1000 \text{ mm}}{1 \text{ m}}$

$a = 2 \times 10^{-4} \times 10^{3} \text{ mm}$

$a = 2 \times 10^{-1} \text{ mm}$

$a = 0.2 \text{ mm}$