Question

the focal length of the objective and eyepiece of an astronomical telescope are 120 cm and 8 cm respectively calculate the magnifying power of astronomical telescope for relaxed eye.​

Answer 1

Answer:

Explanation:

Magnifying power (M)is given by:

Fₐ⇒8 cm

F₀⇒120 cm

So we know that M:-Fₐ/F₀(1+F₀/d)

So to find least distance of distinct vision we need to square Fₐ

     So Fₐ⇒8 cm

               ⇒8 ×8 cm

                ⇒64 cm

       So, d⇒64 cm

So, M⇒-120/8(1+8/64)

        ⇒-15(1+1/8)

         ⇒-40(9/8)

           ⇒-45 cm

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