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. find magnification if placed at the shorter distance find 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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