A point object O is placed on the principle axis of a convex lens of focal length 10 cm at 12 cm from the lens. When object is displaced 1 mm along the principle axis magnitude of displacement of image is x1. When the lens is displaced by 1 mm perpendicular to the principal axis displacement of image is x2 in magnitude. Find the value of x1/x2 .
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For a convex lens when object is placed between focal point and optic centre then the position of the image is …………………
C) same side of the lens
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An object is placed at the following distances from a convex lens of focal length 20 cm. Now match the following.
D) 1 → d, 2 → a, 3 → b, 4 → c
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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 the image distance of a convex lens of focal length 25 cm is 75 cm then the object distance is ……………………
Correct option is A) -37.5 cm
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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 point object O is placed on the principle axis of a convex lens of focal length 10 cm at 12 cm from the lens.
U = -12 f = -10 1/v - 1/u = 1/f 1/v + 1/12 = 1/10 1/v = 1/60 \(-\frac{1}{v^2} \)dv or \(\frac{1}{u^2} \)du du = \(\frac{v^2}{u^2} \)du \( \frac{v}{u} \)= same path △u1 = △u2 = 1mm △u1 = △v2 △v2 = vf - vi x2 = △v2 + 1mm x2...
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