CBSE Class 11-science Answered

 We can use Rayleigh’s to determine the resolving power of telescopr. The angular separation between two objects must be

△θ = 1.22 λ × d

Resolving power = ( 1 / △θ )= d / ( 1.22 λ )

Thus, the higher the diameter d, the better the resolution. The best astronomical optical telescopes have mirror diameters

as large as 10 m to achieve the best resolution.

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Figure shows the position of objective lens and eye piece lens as given in the question.

If the telescope is focussed to an object at a distance 200 cm from objective lens, then the image distance at other side of objective lense is determined from the lens equation as follows

(1/v) - (1/u) = 1/f

where v is image distance , u = 200 cm is object distance and f = 50 cm is focal length of objective lens

Hence we get , (1/ v ) = ( 1/ 50 ) - ( 1/200 ) = ( 3/200 )   or   v = ( 200/3 )  = 66.67 cm 

Image-1 formed by objective lens is used as object for eyepiece lens  to get final image image-2 at distinct vision , i.e. 25 cm from eye piece .

We get distance between eyepiece lens and image-1 from lens equation

(1/v) - (1/u) = 1/f 

-(1/25) + (1/u) = (1/5)  or   1/u = (1/5) + (1/25) = ( 6/25 )   or   u = 4.17 cm

Hence separation distance between objective and eyepiece = ( 66.67 + 4.17) = 70.84 cm

Magnification of image-1 = v/u = 66.67/(-200) = -0.33  

Magnification of image-2 = v/u = -25 / -4.17 ≈ 6 

Hence overall magnification =  (-0.33) × 6  =  -1.98