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

Step 1: Recall the Airy disc radius.
A parallel beam through a circular aperture focused by a lens forms an Airy pattern. The linear radius of the central bright spot on the focal plane is $r = \dfrac{1.22\,\lambda f}{d}$.
Step 2: Convert the data to SI.
Wavelength $\lambda = 500\,nm = 5\times 10^{-7}\,m$, focal length $f = 20\,cm = 0.2\,m$, aperture diameter $d = 5\,cm = 0.05\,m$.
Step 3: Plug into the formula.
$r = \dfrac{1.22\times (5\times 10^{-7})\times 0.2}{0.05}$.
Step 4: Work out the numerator.
$1.22\times 5\times 10^{-7} = 6.1\times 10^{-7}$, and times $0.2$ gives $1.22\times 10^{-7}$.
Step 5: Divide by the diameter.
$r = \dfrac{1.22\times 10^{-7}}{0.05} = 2.44\times 10^{-6}\,m = 2.44\,\mu m$.
Step 6: Conclude.
The radius of the central bright diffraction spot is about $2.44\,\mu m$, which is option (4).
\[ \boxed{r \approx 2.44\,\mu m} \]