Question

The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1st secondary maxima will be :

• A. 2 mm
• B. 2 cm
• C. 0.02 mm
• D. 0.2 mm

Answer 1

The Correct Option is A

The problem asks for the width of the first secondary maxima in a single-slit diffraction pattern. We are given the wavelength of light, the slit width, and the focal length of the convex lens used to focus the pattern.

Concept Used:

In a single-slit diffraction pattern, dark fringes (minima) are observed at angles \(\theta\) that satisfy the condition:

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

where \(a\) is the slit width, \(\lambda\) is the wavelength of light, and \(m = \pm 1, \pm 2, \pm 3, \ldots\) is the order of the minimum.

The secondary maxima are located approximately halfway between the minima.

The width of any secondary maxima is the distance between the two minima that bound it. The first secondary maxima is located between the first minimum (\(m=1\)) and the second minimum (\(m=2\)).

For small angles (which is a valid approximation in most diffraction experiments), \(\sin(\theta) \approx \tan(\theta) \approx \theta\). If the pattern is focused on a screen at a distance \(D\) from the slit (here, \(D\) is the focal length \(f\) of the lens), the position \(y\) of a fringe from the center is given by \(y = D \tan(\theta) \approx D\theta\).

So, the position of the \(m^{th}\) minimum is given by:

\[ y_m = \frac{m\lambda D}{a} \]

The width of the first secondary maxima (\(W\)) is the difference in position between the 2nd and 1st minima:

\[ W = y_2 - y_1 \]

Step-by-Step Solution:

Step 1: List the given values and convert them to SI units.

• Wavelength of light, \(\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}\).
• Slit width, \(a = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m}\).
• Focal length of the lens (distance to the screen), \(D = f = 100 \, \text{cm} = 1 \, \text{m}\).

Step 2: Find the position of the first minimum (\(y_1\)).

Using the formula for the position of the \(m^{th}\) minimum with \(m=1\):

\[ y_1 = \frac{1 \cdot \lambda D}{a} \] \[ y_1 = \frac{(400 \times 10^{-9} \, \text{m}) \times (1 \, \text{m})}{0.2 \times 10^{-3} \, \text{m}} \] \[ y_1 = \frac{400}{0.2} \times 10^{-6} \, \text{m} = 2000 \times 10^{-6} \, \text{m} = 2 \times 10^{-3} \, \text{m} = 2 \, \text{mm} \]

Step 3: Find the position of the second minimum (\(y_2\)).

Using the formula for the position of the \(m^{th}\) minimum with \(m=2\):

\[ y_2 = \frac{2 \cdot \lambda D}{a} \] \[ y_2 = \frac{2 \times (400 \times 10^{-9} \, \text{m}) \times (1 \, \text{m})}{0.2 \times 10^{-3} \, \text{m}} \] \[ y_2 = 2 \times y_1 = 2 \times (2 \times 10^{-3} \, \text{m}) = 4 \times 10^{-3} \, \text{m} = 4 \, \text{mm} \]

Step 4: Calculate the width of the first secondary maxima.

The first secondary maxima lies between the first and second minima. Its width is the distance between these two minima.

\[ W = y_2 - y_1 \] \[ W = 4 \, \text{mm} - 2 \, \text{mm} = 2 \, \text{mm} \]

Final Computation & Result:

The width of the first secondary maxima is the distance between the first and second dark fringes. Alternatively, the angular width of any secondary maxima is \( \Delta\theta = \frac{\lambda}{a} \). The linear width is \( W = D \Delta\theta = \frac{D\lambda}{a} \).

Using this direct formula:

\[ W = \frac{(1 \, \text{m}) \times (400 \times 10^{-9} \, \text{m})}{0.2 \times 10^{-3} \, \text{m}} \] \[ W = \frac{400}{0.2} \times 10^{-6} \, \text{m} = 2000 \times 10^{-6} \, \text{m} = 2 \times 10^{-3} \, \text{m} \] \[ W = 2 \, \text{mm} \]

The width of the 1st secondary maxima is 2 mm.

Answer 2

The width of the first secondary maxima for single-slit diffraction is given by:

Width of 1^st secondary maxima = \( \frac{\lambda D}{a} \)

where \( \lambda = 400 \times 10^{-9} \, \text{m} \), \( a = 0.2 \times 10^{-3} \, \text{m} \), and \( D = 100 \, \text{cm} = 1 \, \text{m} \). Substitute the values:

Width of 1^st secondary maxima = \[ \frac{400 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 2 \, \text{mm} \]