Question

A parallel beam of monochromatic light of wavelength 5000 •Å is incident normally on a single narrow slit of width 0.001 mm. The light is focused by convex lens on screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of ________ (degree).

Answer 1

Correct Answer: 30

To determine the angle of diffraction for the first minima in a single slit diffraction pattern, we use the formula for minima:
a*sin(θ) = m*λ,
where a = slit width, θ = angle of diffraction, m = order of minima (for the first minima, m=1), and λ = wavelength of light.

Given: 
a = 0.001 mm = 1 x 10^-6 m,
λ = 5000 Å = 5000 x 10^-10 m.

Using the formula for the first minima (m=1):
1 x 10^-6 * sin(θ) = 1 * 5000 x 10^-10.
Solving for sin(θ):
sin(θ) = (5000 x 10^-10) / (1 x 10^-6) = 0.5.

Now, calculate θ:
θ = sin^-1(0.5) = 30°.

Therefore, the angle of diffraction for the first minima is 30°.

Answer 2

For the first minima in a single-slit diffraction pattern:

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

Given:
- \( a = 0.001 \, \text{mm} = 1 \times 10^{-6} \, \text{m} \),
- \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \).

Substituting:

\(\sin \theta = \frac{\lambda}{a} = \frac{5000 \times 10^{-10}}{1 \times 10^{-6}} = 0.5\)

\(\theta = \sin^{-1}(0.5) = 30^\circ\)
The Correct answer is: 30