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Geometrical Optics question

2022 · 28 Jun · Shift 1 · Q55

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Geometrical Optics
2022 · 28 Jun · Shift 1 · Q55

Geometrical Optics question

2022 · 28 Jun · Shift 1 · Q55

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1

The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440

\mathop A\limits^o

is used to see th object will be :

A
8.1 $\times$ 106
B
10.0 $\times$ 107
C
8.2 $\times$ 105
D
1.0 $\times$ 10 $-$ 8

View written solutionFree

Correct answer: C

Use Rayleigh criterion for telescope

For a telescope, the minimum angular resolution is

\theta_{\min} = 1.22\frac{\lambda}{D}

where:

$\lambda$ = wavelength of light
$D$ = aperture diameter of objective

The resolving power of a telescope is

\text{Resolving power} = \frac{1}{\theta_{\min}} = \frac{D}{1.22\lambda}

Convert given quantities into SI units

Aperture:

D = 24.4\text{ cm} = 0.244\text{ m}

Wavelength:

\lambda = 2440\,\text{\AA} = 2440 \times 10^{-10}\text{ m} = 2.44 \times 10^{-7}\text{ m}

Substitute into the formula

\text{Resolving power} = \frac{0.244}{1.22 \times 2.44 \times 10^{-7}}

Since

1.22 \times 2.44 = 2.9768

so,

\text{Resolving power} = \frac{0.244}{2.9768 \times 10^{-7}}

= \left(\frac{0.244}{2.9768}\right) \times 10^7

\approx 0.08195 \times 10^7

\approx 8.2 \times 10^5

Match with the options

Thus, the resolving power is

\boxed{8.2 \times 10^5}

So the correct option is C.

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