Wheh the object is at distances x1 and x2 from the thin lens, theh the

1.	Wheh the object is at distances x1 and x2 from the thin lens, theh the object is at distances x and 2 from the thin lens, the two images formed areof same size but the one is real and the other is virtual, respectively. The focal length cofthe lens is
Answer» Let us say it is a convex lens of focal length f. It can form both real and virtual images. 1/v1 - 1/x= 1/f, here v & f are positive and u is negative.1/v1 + 1/x1 = 1/f --- (1)=> x1/v1 + 1 = x1/f -- (2) The virtual image is formed in front of the lens. So v and u1 are negative.- 1/v2 + 1/x2 = 1/f --- (3)- x2/v2 + 1 = x2/f --(4) Magnification = v1/x1 = v2/x2 ---(5) Add (2) and (4) and use (5) to get: 2 = x1/f + x2/f => f = (x1 + x2)/2

Wheh the object is at distances x1 and x2 from the thin lens, theh the object is at distances x and 2 from the thin lens, the two images formed areof same size but the one is real and the other is virtual, respectively. The focal length cofthe lens is

Answer»

Let us say it is a convex lens of focal length f. It can form both real and virtual images.

1/v1 - 1/x= 1/f, here v & f are positive and u is negative.1/v1 + 1/x1 = 1/f --- (1)=> x1/v1 + 1 = x1/f -- (2)

The virtual image is formed in front of the lens. So v and u1 are negative.- 1/v2 + 1/x2 = 1/f --- (3)- x2/v2 + 1 = x2/f --(4)

Magnification = v1/x1 = v2/x2 ---(5)

Add (2) and (4) and use (5) to get: 2 = x1/f + x2/f => f = (x1 + x2)/2

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Wheh the object is at distances x1 and x2 from the thin lens, theh the object is at distances x and 2 from the thin lens, the two images formed areof same size but the one is real and the other is virtual, respectively. The focal length cofthe lens is

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