Two stars are 10 light years away from the earth. They are seen through a telescope of objective diameter 30 cm. The wavelength of light is 600 nm. To see the stars just resolved by the telescope, the minimum distance between them should be ( 1 light year = 9.46 × 1015 m) of the order of :

Question

Two stars are 10 light years away from the earth. They are seen through a telescope of objective diameter 30 cm. The wavelength of light is 600 nm. To see the stars just resolved by the telescope, the minimum distance between them should be ( 1 light year = 9.46 × 1015 m) of the order of : (A) 106 km (B) 108 km (C) 1011 km (D) 1010 km

Tardigrade Admin · Accepted Answer

We have, Δ θ=(0.61 λ/9)=(1/ R ) 1=( R /9) 0.61 × λ =(9.46 × 1015 × 10 × 0.61 × 600 × 10-9/0.3) =1.15 × 1011 m =1.115 × 108 km

$1 0^{6} km$

$1 0^{8} km$

$1 0^{11} km$

$1 0^{10} km$

We have, $Δ θ = \frac{0.61 λ}{9} = \frac{1}{R}$
$1 = \frac{R}{9} 0.61 \times λ$
$= \frac{9.46 \times 1 0 ^{15} \times 10 \times 0.61 \times 600 \times 1 0 ^{- 9}}{0.3}$
$= 1.15 \times 1 0^{11} m$
$= 1.115 \times 1 0^{8} km$

Q. Two stars are 10 light years away from the earth. They are seen through a telescope of objective diameter 30cm. The wavelength of light is 600nm. To see the stars just resolved by the telescope, the minimum distance between them should be ( 1 light year =9.46×1015m) of the order of :

106km

108km

1011km

1010km

We have, Δθ=90.61λ​=R1​ 1=9R​0.61×λ =0.39.46×1015×10×0.61×600×10−9​ =1.15×1011m =1.115×108km

$1 0^{6} km$

$1 0^{8} km$

$1 0^{11} km$

$1 0^{10} km$

We have, $Δ θ = \frac{0.61 λ}{9} = \frac{1}{R}$
$1 = \frac{R}{9} 0.61 \times λ$
$= \frac{9.46 \times 1 0 ^{15} \times 10 \times 0.61 \times 600 \times 1 0 ^{- 9}}{0.3}$
$= 1.15 \times 1 0^{11} m$
$= 1.115 \times 1 0^{8} km$