Question

A monochromatic light of wavelength 6000Å is incident on the single slit of width 0.01 mm. If the diffraction pattern is formed at the focus of the convex lens of focal length 20 cm, the linear width of the central maximum is :

• A. 60 mm
• B. 24 mm
• C. 120 mm
• D. 12 mm

Answer 1

The Correct Option is B

The linear width of the central maximum is given by the formula:

\[ W = \frac{2\lambda D}{a} \]

where:

• \(\lambda = 6000 \, \text{\AA} = 6 \times 10^{-7} \, \text{m}\)
• \(D = 20 \, \text{cm} = 0.2 \, \text{m}\)
• \(a = 0.01 \, \text{mm} = 1 \times 10^{-5} \, \text{m}\)

Substituting the values:

\[ W = \frac{2 \times 6 \times 10^{-7} \times 0.2}{1 \times 10^{-5}} = 24 \, \text{mm} \]

Thus, the correct answer is Option (2).

Answer 2

A monochromatic light of wavelength 6000 Å is incident on a single slit of width 0.01 mm. The diffraction pattern is formed at the focus of a convex lens of focal length 20 cm. We need to find the linear width of the central maximum.

Concept Used:

In single-slit Fraunhofer diffraction, the angular width of the central maximum is the angle between the first minima on either side of the central maximum. The angular position of the first minimum is given by:

\[ \sin \theta \approx \theta = \frac{\lambda}{a} \]

where \( \lambda \) is the wavelength of light and \( a \) is the slit width. The angular width of the central maximum is \( 2\theta \). The linear width on the screen placed at the focal plane of the lens is given by:

\[ \text{Linear width} = 2 \theta \times f = 2 \frac{\lambda}{a} \times f \]

where \( f \) is the focal length of the lens.

Step-by-Step Solution:

Step 1: Convert all given quantities to consistent units (meters).

\[ \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \] \[ a = 0.01 \, \text{mm} = 0.01 \times 10^{-3} \, \text{m} = 1 \times 10^{-5} \, \text{m} \] \[ f = 20 \, \text{cm} = 0.2 \, \text{m} \]

Step 2: Write the formula for the linear width of the central maximum.

\[ \text{Linear width} = 2 \frac{\lambda}{a} \times f \]

Step 3: Substitute the values into the formula.

\[ \text{Linear width} = 2 \times \frac{6 \times 10^{-7}}{1 \times 10^{-5}} \times 0.2 \]

Step 4: Simplify the expression step by step.

\[ \frac{6 \times 10^{-7}}{1 \times 10^{-5}} = 6 \times 10^{-2} = 0.06 \] \[ 2 \times 0.06 = 0.12 \] \[ 0.12 \times 0.2 = 0.024 \, \text{m} \]

Step 5: Convert the result to millimeters.

\[ 0.024 \, \text{m} = 24 \, \text{mm} \]

Thus, the linear width of the central maximum is 24 mm.