Question

If a parallel beam of light of wavelength $500\text{ nm}$ is incident on a convex lens of focal length $20\text{ cm}$ having a circular aperture of diameter $5\text{ cm}$, then the radius of the central bright diffraction spot formed on the focal plane of the lens is nearly (in $\mu\text{m}$):

• A. $1.83$
• B. $0.61$
• C. $1.22$
• D. $2.44$

Hint:

For a circular aperture, the radius of the Airy disc formed at the focal plane of a lens is \[ r=\frac{1.22\lambda f}{d}. \] Remember that:

• Radius of Airy disc $\propto$ wavelength.

• Radius of Airy disc $\propto$ focal length.

• Radius of Airy disc $\propto \frac{1}{d}$.
A larger aperture produces a smaller diffraction spot and hence better resolving power.

Answer 1

The Correct Option is D

Concept: When a parallel beam of monochromatic light passes through a circular aperture and is brought to focus by a convex lens, diffraction takes place due to the wave nature of light. Instead of forming a perfect point image, a diffraction pattern known as the

Airy pattern is produced on the focal plane of the lens. The central bright circular region of this diffraction pattern is called the

Airy Disc. The boundary of this bright region is determined by the position of the first diffraction minimum. For a circular aperture, the angular radius of the first minimum is given by Airy's criterion: \[ \theta=\frac{1.22\lambda}{d} \] where \[ \lambda=\text{wavelength of light} \] and \[ d=\text{diameter of the circular aperture}. \] If the diffraction pattern is observed on the focal plane of a lens of focal length $f$, then the linear radius $r$ of the central bright spot is \[ r=f\theta. \] Combining the two relations, we obtain \[ r=\frac{1.22\lambda f}{d}. \] This formula directly gives the radius of the Airy disc formed on the focal plane.

Step 1: Convert all given quantities into SI units. The wavelength of light is \[ \lambda=500\text{ nm}. \] Since \[ 1\text{ nm}=10^{-9}\text{ m}, \] therefore \[ \lambda=500\times10^{-9}\text{ m} \] or \[ \lambda=5\times10^{-7}\text{ m}. \] The focal length of the lens is \[ f=20\text{ cm}. \] Since \[ 1\text{ cm}=10^{-2}\text{ m}, \] we get \[ f=20\times10^{-2}=0.2\text{ m}. \] The diameter of the aperture is \[ d=5\text{ cm} \] which becomes \[ d=5\times10^{-2}\text{ m} \] or \[ d=0.05\text{ m}. \] Thus, \[ \lambda=5\times10^{-7}\text{ m}, \] \[ f=0.2\text{ m}, \] \[ d=0.05\text{ m}. \]

Step 2: Apply the Airy disc radius formula. The radius of the central bright diffraction spot is \[ r=\frac{1.22\lambda f}{d}. \] Substituting the given values, \[ r= \frac{1.22\times(5\times10^{-7})\times0.2} {5\times10^{-2}}. \]

Step 3: Simplify the numerical expression carefully. First cancel the factor $5$ present in the numerator and denominator: \[ r= 1.22\times \frac{10^{-7}\times0.2} {10^{-2}}. \] Since \[ \frac{10^{-7}}{10^{-2}} = 10^{-5}, \] therefore \[ r = 1.22\times0.2\times10^{-5}. \] Multiplying, \[ r = 0.244\times10^{-5}. \] Rewriting, \[ r = 2.44\times10^{-6}\text{ m}. \]

Step 4: Convert the answer into micrometres. We know that \[ 1\,\mu\text{m}=10^{-6}\text{ m}. \] Hence, \[ 2.44\times10^{-6}\text{ m} = 2.44\,\mu\text{m}. \] Therefore, \[ r=2.44\,\mu\text{m}. \]

Final Conclusion: The radius of the central bright diffraction spot (Airy disc) formed on the focal plane of the lens is \[ \boxed{2.44\,\mu\text{m}}. \] Hence, the correct answer is \[ \boxed{\text{(D) }2.44}. \]