(a) How many astronomical units (A.U) make 1 parsec ? (b) Consider a sunlike star at a distance of 1 parsec.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 can't resolve object smallar then 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 A.U. from the erath. Calculate what size it will appear when seen through the same telescope.

Explanation

(a) By definition, 1 par sec= distance at which 1 A.U. long arc subtends an angle of 1 sec. `1 par sec= (1 A.U)/(1 arc sec)`
As `1 arc sec =(1)/(60xx60)xx(pi)/(180) radian`
`:. 1 par sec= (1 A.U)/(pi//60xx60xx180) = (3600xx180)/(pi) A.U. = 206181 A.U.`
` 1 par sec = 2xx10^5 A.U.`
(b) At ` 1 A.U., sun's "diameter" is (1^@)/(2).`
`:. At 1 par sec = 2xx10^5 A.U., star's "diameter will be" = (1//2^@)/(2xx10^5) = (60')/(4xx10^5) = 15xx10^(-5)min`
With magnification = 100 star look of diameter `=15xx10^(-5)xx 100 = (15xx10^9-3)min.`
However, due to atmospheric fultuations, it will still look of diameter about 1 minute.
Stars can't be magnified using telescope.
(c ) we are given `(D_(mars))/(D_(earth)) =(1)/(2)` , whrer D is for diameter.
From Ques. 2.25 (c), we know = `(D_(earth))/(D_(sun)) = (1)/(100) :. (D_(mars))/(D_(sun)) = (1)/(2)xx(1)/(100) = (1)/(200)`
At 1 A.U. sun's diameter = `(1^@)/(2)` :. mar's diameter = `(1)/(2)xx(1)/(100) = (1^@)/(400)`
At `(1)/(2) A.U`., mar's diameter = `(1)/(400)xx2^@ = (1^@)/(200)`.
with magnification = 100, mar's diameter `= (1)/(200)xx100^@ = (1^@)/(2) = 30`'
This is larger then resolution limit due to atmopheric fluctuations. Hence, we conclude that planet looks magnified when seen through telescope.