Enter question here... A transparent photographic slide is placed in front of th

Question

Enter question here...

A transparent photographic slide is placed in front of the converging lens with a focal length of 2.27 cm. An image of the slide is formed 14.2 cm from the slide. How far is the lens from the slide if the image is real?( Enter your answers from smallest to largest starting with the first answer blank. Enter NONE in any remaining answer blanks.) cm cm How far is the lens from the slide if the image is virtual? (Enter your answers from smallest to the largest starting with the first answer blank. Enter NONE in any remaining answer blanks.) cm cm

Explanation / Answer

Looks like you are answering a homework or exam question so don't cheat. But you can use the formula 1/u +1/v=1/f. Then you will get a quadratic equation you have to solve - just boring arithmetic. The tricky part is remembering that a virtual image has a minus distance from the lens. And to confuse you the question expresses the image distance from the object not from the lens. But if you try to figure it out yourself you might learn something. u = object distance, v = image distance (v always positive). Real is positive sign convention. (a) 1 / u + 1 / v = 1 / 2.33 u + v = 11.4 Eliminating v: 1 / u + 1 / (11.4 - u) = 1 / 2.33 11.4/ [ u(11.4 - u) ] = 1 / 2.33 2.33 * 11.4 = 11.4u - u^2 u^2 - 11.4u + 26.562 = 0 u = [ 13.5 +/- sqrt(13.5^2 - 4 * 33.075) ] / 2 = 8.135 cm or 3.265 cm. (b) 1 / u - 1 / v = 1 / 2.33 v - u = 11.4 1 / u - 1 / (11.4 + u) = 1 / 2.33 11.4 / [ u(11.4 + u) ] = 1 / 2.33 2.33 * 11.4 = 11.4u + u^2 u^2 + 11.4u - 26.562 = 0 u = [ - 13.5 +/- sqrt(13.5^2 + 4 * 33.075) ] / 2 The only positive root is u = 1.98 cm.

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