A convex lens has focal length of 2.0 cm. Find its magnifying power if image is formed at DDV.
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When a lens of focal length f is cut in to two equal halves perpendicular to principal axis, then the focal length of each part of the lens is ………………
Correct option is B) double
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If 40 cm each is the object and image distance respectively for a convex lens then the focal length is ……………………
Correct option is C) 20 cm
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if a convex lens has its ob j ect and image distance equal (say x) the focal length is equal to …………………
Correct option is B) x/2
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An image is formed at distance of 30 cm from the centre of convex lens of focal length 15 cm. The distance of the object is ………………..
Correct option is C) 30 cm
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A convex lens of focal length 20 cm is cut into two halves along principal axis then focal length of each part will be …………………
Correct option is A) 20 cm
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A convex lens of focal length 20 cm is cut into two halves perpendicular to the principal axis then focal length of each part will be ……………….
Correct option is C) 40 cm
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What are convex and concave lenses? For which condition, convex lens will have negative focal length?
1. A lens is said to be convex if it is thicker in the middle and narrowing towards the periphery. According to Cartesian sign convention, its focal length is positive. 2....
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A magnifying glass of focal length 10 cm is used to read letters of thickness 0.5 mm held 8 cm away from the lens. Calculate the image size.
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...
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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
Given: M.P = 12, v = D, f0 = 90 cm, To find: Focal length of eye piece (fe) Formula: M.P = \(\frac{f_0}{f_e}\) (1 + \(\frac{f_e}{D}\)) Calculation: From formula. 12 = \(\frac{90}{f_e}\) (1 + \(\frac{f_e}{25}\)) ∴ fe = 10.71 cm
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Two convex lenses of an astronomical telescope have focal length 1.3 m and 0.05 m respectively. Find the magnifying power and the length
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...
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