QInteger TypeOptics
A distant star is to be observed by a telescope at an angular resolution of $3.0\times10^{-7}$ radian. If the wavelength of light from the star reaching the telescope is 500 nm, the minimum diameter of the objective lens of the telescope is ________ cm (nearest integer).
✅ Correct Answer
203 cm
Solution
1
Apply Rayleigh's Criterion for angular resolution

The minimum angular resolution (limit of resolution) for a circular aperture of diameter $a$ is given by:

$$\theta = \frac{1.22\,\lambda}{a}$$

where $\theta$ is the minimum angle that can be resolved, $\lambda$ is the wavelength, and $a$ is the aperture diameter.

2
Substitute the given values
$\theta = 3.0\times10^{-7}$ rad, $\lambda = 500$ nm $= 500\times10^{-9}$ m

$a = \dfrac{1.22\,\lambda}{\theta} = \dfrac{1.22\times500\times10^{-9}}{3.0\times10^{-7}}$

$= \dfrac{610\times10^{-9}}{3.0\times10^{-7}} = \dfrac{610}{300}$ m $\approx 2.033$ m
3
Convert to centimetres
$a = 2.033\ \text{m} = 203.3\ \text{cm} \approx \boxed{203}\ \text{cm}$
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Theory & Key Concepts
Rayleigh's Criterion and Resolving Power

Rayleigh's Criterion states that two point sources are just resolved when the central maximum of the diffraction pattern of one source falls on the first minimum of the other. For a circular aperture of diameter $a$, the angular limit of resolution is:

$\theta_{min} = \dfrac{1.22\lambda}{a}$

Resolving power is defined as $1/\theta_{min}$. A larger aperture gives a smaller $\theta_{min}$, meaning it can resolve finer details — this is why larger telescopes see more detail.

Why 1.22? For a circular aperture (unlike a slit), the first diffraction minimum occurs at $\sin\theta = 1.22\lambda/a$. The factor 1.22 comes from the first zero of the Bessel function $J_1(x)$, which describes circular diffraction.

Practical significance: The Hubble Space Telescope has a 2.4m mirror and achieves $\theta \approx 0.05$ arcseconds. Ground-based telescopes are limited by atmospheric turbulence, not diffraction. Modern techniques like adaptive optics correct for atmospheric distortion.

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Important Concepts
Rayleigh Criterion Formula

$\theta_{min}=1.22\lambda/a$. For this problem: $a=1.22\times500\times10^{-9}/(3\times10^{-7})=2.033$ m. The factor 1.22 arises from circular aperture diffraction (Bessel function first zero).

Why Larger Telescope = Better Resolution

Resolving power $= 1/\theta_{min} = a/(1.22\lambda)$. A larger diameter $a$ gives smaller minimum angle, so you can distinguish finer detail. This is why observatories use large mirrors.

Wavelength Dependence

Shorter wavelength → smaller diffraction limit → better resolution. UV telescopes can resolve finer detail than optical telescopes of the same aperture. Radio telescopes need enormous dishes because radio wavelengths are millions of times longer.

Units Reminder

Always convert to SI first: $\lambda = 500$ nm $= 5\times10^{-7}$ m. Answer $2.033$ m $\times 100 = 203.3$ cm, nearest integer $= 203$ cm.

FAQs
1
What is Rayleigh's criterion?
Two objects are just resolved when the central maximum of one's diffraction pattern falls on the first minimum of the other: $\theta_{min}=1.22\lambda/a$.
2
Why 1.22 and not 1?
For a circular aperture, the first minimum of the Airy disc (circular diffraction pattern) is at $1.22\lambda/a$, not $\lambda/a$. The 1.22 factor comes from the first root of the Bessel function $J_1$.
3
What is 500 nm in metres?
$500\text{ nm}=500\times10^{-9}\text{ m}=5\times10^{-7}\text{ m}$.
4
Why convert answer to cm?
Question asks for answer in cm as nearest integer. $2.033\text{ m}=203.3\text{ cm}\approx203\text{ cm}$.
5
Is this from JEE Main 2026?
Yes, this appeared in JEE Main 2026 Physics Section B.
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