Magnifying power of an astronomical telescope is 12 and the image is formed at D.D.V. If the focal length of the objective is 90 cm

1. Telescopes used to see the objects on the Earth, like mountains, trees, players playing a match in a stadium, etc. are called terrestrial telescopes.  2. In such case, the final...

Correct option is  B) double

Correct option is  C) 30 cm

Correct option is  A) 20 cm

Correct option is  C) 40 cm

Given: f = 2 cm, v = D = 25 cm To find: Magnifying power (M.P.) Formula: M.P = 1 + \(\frac{D}{f}\) Calculation: From formula, M.P = 1 + \(\frac{25}{2}\) M.P. = 1 + 12.5 = 13.5

Given that, f = +10 cm, u = -8 cm, From thin lens formula, \(\frac{1}{f}=\frac{1}{\text{v}}-\frac{1}{u}\) ∴ \(\frac{1}{10}=\frac{1}{\text{v}}-\frac{1}{-8}\) ∴ v = 40 cm Magnification of a lens is, m = \(\frac{v}{u}=\frac{Object\,size\,h-i}{Object\,size\,h-0}\) ∴ \(\frac{40}{8}=\frac{h_1}{0.5}\) ∴ h1 = 2.5 cm This implies the height of the...

Given: f0 = 1 m, L = 1.05 m To find: i. Focal length of eyepiece (fe) Magnifying power under normal adjustment (M) Formula: i. L = f0 + fe ii. M = \(\frac{f_0}{f_e}\) Calculation: From formula (i), fe = L – f0 =...

Given: f0 = 1.3 m, fe = 0.05 m To find: i. Magnifying power of telescope (M.P.) ii. Length of telescope (L) Formulae: i. M.P = \(\frac{f_0}{f_e}\) ii. L = f0 + fe Calculation: From formula (i), M.P = \(\frac{1.3}{0.05}\) = 26 From formula (ii), L...

 A) \(\cfrac{(n+1)f}n\)