A high school astronomy teacher wants to build a refracting telescope to view th

Question

A high school astronomy teacher wants to build a refracting telescope to view the surface features of Mars. If an angular magnification of -50.0 is desired and an eyepiece of focal length 2.50 cm is to be used, what must be the focal length of the objective lens? She wants to resolve Martian surface features of diameter 115 km when Mars is 2.01 × 108 km from Earth. If the average wavelength of light used with this telescope will be 549 nm, what is the minimum diameter of the objective lens needed to acheive this resolution?

Explanation / Answer

M = -50 ; fe = 2.5 cm ; fo = ? ; D = 115 km ; d = 2.01 x 10^8 km ; lambda = 549 nm

The angle that will be substended at the objective lens would be:

theta = D/d = 1.15 x 10^2 km/2.01 x 10^8 km = 0.572 x 10^-6 rad

We know that,

theta = (1.22) lambda/a ; where a is the diameter of the objective

a = 1.22 lambda/ theta

a = 1.22 x 549 x 10^-9/0.572 x 10^-6 = 1.17 m

Hence, a = 1.17 m

we know that,

m = -fo/fe

-50 = -fo/2.5

fo = 50x2.5 = 125 cm

Hence, fo = 125 cm

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