**Question:**

(a) How many astronomical units (AU) make 1 parsec?

(b) Consider the sun like a star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the

angular size of the star? Sun appears to be (1/2)° from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than

1 arc minute.

(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 AU from the earth. Calculate at what

size it will appear when seen through the same telescope.

**Solution:**

(a) From the definition, 1 parsec is equal to the distance at which 1 AU long arc subtends an angle of 1s.

Using the definition, we can say that

1 parsec = (3600)(180)/π AU

= 206265 AU

= 2 × 105 AU

b) Given that the sun’s angular diameter from the earth is 1/2 degree at 1 AU.

Angular diameter of the sun like star at a distance of 2 parsec

= [(1/2)/(2)(2)(105)] degree

= 1/8 × 10-5 degree

= 7.5 × 10-5 arcmin

When the sun appears like a start through the telescope that has a magnification of 100, the angular diameter of the star is

= (100)(7.5 × 10-5)

= 7.5 × 10-3 arcmin

But the angular size of the sun appears as 1 arcmin to the eyes as the eyes cannot resolve smaller than 1 arcmin because of atmospheric fluctuations.

c) Given that,

Dmars/Dearth = 1/2

We also know that Dearth/Dsun = 1/100

Therefore, Dmars/Dsun = 1/2 × 1/100

At 1AU, the sun’s diameter = (1/2) degree

Therefore, diameter of mars = (1/400) degree

At 1/2 AU, mars diameter = (1/400)(2) = (1/200) degree

With 100 magnification, mars diameter = (1/2) degree = 30’

Therefore, it can be said that the value is larger than the resolution limit because of atmospheric fluctuations and hence it looks magnified.

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