An object is placed 69.0 cm from a screen. (a) Where should a converging lens of

Question

An object is placed 69.0 cm from a screen. (a) Where should a converging lens of focal length 9.0 cm be placed to form an image on the screen? shorter distance_________cm from the screen farther distance_________cm from the screen (b) Find the magnification of the lens. magnification if placed at the shorter distance magnification if placed at the farther distance

Explanation / Answer

1/f = 1/u + 1/v distance from screen and object = D = 69cm = u + v 1/9 = uv/u +v 1/9 = uv/ 69 uv = 69/9 uv = 23/3 and u + v = 69 we get a quadratic equation 3u^2 - 207u + 23 = 0 ; u = 68.89 cm and 0.11 cm V = 0.11 cm and 68.89cm . Hence shorter distance = 0.11cm and farther distance = 68.89cm b) magnification at shorter distance => v=0.11cm and u = 68.89 cm ; M = v/u = 1.59 * 10^(-3) magnification at longer distance => v = 68.89 cm and u = 0.11 cm M = v/u = 626.27

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