A linear aperture whose width is $0.02\, cm$ is placed immediately in front of a lens of focal length $60\, cm$. The aperture is illuminated normally by a parallel beam of wavelength $5 \times 10^{-5} cm$. The distance of the first dark band of the diffraction pattern from the centre of the screen is

Question

What is the frequency of a wave with a wavelength of $ 2 \, \text{m} $ and a velocity of $ 4 \, \text{m/s} $?

Answer 1

To solve this problem, we need to determine the distance of the first dark band (or first minimum) of the diffraction pattern formed by a single slit. The relevant principle governing this is the single-slit diffraction pattern, where the position of the first minimum is given by the formula:

a \sin \theta = m \lambda

where:

a is the width of the slit.
\theta is the angle at which the first minimum occurs.
m is the order of the minimum (for first minimum, m = 1).
\lambda is the wavelength of the light.

Given values:

Width of the aperture, a = 0.02 \, \text{cm}
Focal length of the lens, f = 60 \, \text{cm}
Wavelength of light, \lambda = 5 \times 10^{-5} \, \text{cm}

For small angles in radians, \sin \theta \approx \theta. Thus, the distance y of the first minimum on the screen from the center is calculated by the formula:

y = f \theta

Substituting the values, the angle \theta can be determined from:

\theta = \frac{m \lambda}{a}

For the first minimum, m = 1

\theta = \frac{1 \times 5 \times 10^{-5}}{0.02}

\theta = \frac{5 \times 10^{-5}}{2 \times 10^{-2}}

\theta = 2.5 \times 10^{-3} \, \text{radians}

Now, calculate the distance y:

y = 60 \times 2.5 \times 10^{-3}

y = 0.15 \, \text{cm}

Therefore, the distance of the first dark band from the center of the screen is 0.15 \, \text{cm}.

This matches the given correct answer option: 0.15 cm.

Answer 2