A telescope is used to resolve two stars separated by 3.2 × 10-6 rad. If the wavelength of light used in 5600 mathringA, what should be the aperture of the objective of the telescope?

Question

A telescope is used to resolve two stars separated by 3.2 × 10-6 rad. If the wavelength of light used in 5600 mathringA, what should be the aperture of the objective of the telescope? (A) 0.2135 m (B) 0.1488 m (C) 0.567 m (D) 1 m

Tardigrade Admin · Accepted Answer

Aperture of the telescope D=(1.22 λ/d θ) Here λ=5600 mathringA=5600 × 10-10 m , d θ=3.2 × 10-6 rad ∴ D=(1.22 × 5600 × 10-10/3.2 × 10-6) ⇒ D=0.2135 m

Q. A telescope is used to resolve two stars separated by $3.2 \times 1 0^{- 6}$ rad. If the wavelength of light used in $5600 \overset{˚}{A}$ , what should be the aperture of the objective of the telescope?

0.2135 m

0.1488 m

0.567 m

1 m

Aperture of the telescope
$D = \frac{1.22 λ}{d θ}$
Here $λ = 5600 \overset{˚}{A} = 5600 \times 1 0^{- 10} m, d θ = 3.2 \times 1 0^{- 6} r a d$
$∴ D = \frac{1.22 \times 5600 \times 1 0 ^{- 10}}{3.2 \times 1 0 ^{- 6}}$
$\Rightarrow D = 0.2135 m$

Q. A telescope is used to resolve two stars separated by 3.2×10−6 rad. If the wavelength of light used in 5600A˚, what should be the aperture of the objective of the telescope?

0.2135 m

0.1488 m

0.567 m

1 m

Aperture of the telescope D=dθ1.22λ​ Here λ=5600A˚=5600×10−10m,dθ=3.2×10−6rad ∴D=3.2×10−61.22×5600×10−10​ ⇒D=0.2135m

Q. A telescope is used to resolve two stars separated by $3.2 \times 1 0^{- 6}$ rad. If the wavelength of light used in $5600 \overset{˚}{A}$ , what should be the aperture of the objective of the telescope?

Aperture of the telescope
$D = \frac{1.22 λ}{d θ}$
Here $λ = 5600 \overset{˚}{A} = 5600 \times 1 0^{- 10} m, d θ = 3.2 \times 1 0^{- 6} r a d$
$∴ D = \frac{1.22 \times 5600 \times 1 0 ^{- 10}}{3.2 \times 1 0 ^{- 6}}$
$\Rightarrow D = 0.2135 m$