A hobby telescope has an objective lens with a focal length of 75 cm and an eyep

Question

A hobby telescope has an objective lens with a focal length of 75 cm and an eyepiece with a focal length of 8.2 mm. We view the planet Jupiter with this telescope. We do this at the time of year when we are closest to Jupiter, a distance of 6.28 × 1011 m. The diameter of Jupiter is 1.40 × 108 m.

(a) Find the diameter (mm) of the image of Jupiter produced by the objective lens.

(b) How far (cm) should I place a marble (1-cm diameter) from my eye so that it appears to be the same size as Jupiter as viewed through the eyepiece of the telescope? Give the answers to 2 significant figures.

THE ANSWER FOR PART A IS 0.17 MM

THE ANSWER FOR PART B IS 49 CM.

COULD ANYONE SHOW ME HOW TO GET THOSE ANSWERS?

Explanation / Answer

(a)    diameter of image / diameter ofJupiter = focal length / distance to Jupiter

.

         diameter ofimage / 1.40 x 109   = 0.75 / 6.28 x 1011

.

         diameter ofimage =   0.001761 meter =    1.761 mm =    1.7 mm

.

(b)    the angular magnification is  f1/f2 = 75 / 0.82 = 91.4

    The angular diameter of Jupiteris         diam of Jupiter/ distance to Jupiter

.

    Angular diameter of imageis      ang mag * diam of Jupiter /distance to Jupiter

.

    Angular diameter of marbleis     diameter of marble / distance tomarble

.

The latter two are equal, so...

.

     diam of marble / distance tomarble = ang mag * diam of Jupiter / distance toJupiter

.

       0.01 / distto marble = 91.4 * 1.40 x 109 / 6.82 x1011

.

         0.01 / distto marble   =    0.1749

.

        dist tomarble   =    0.049 m = 49 cm

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