A transparent photographic slide is placed in front of a converging lens with a

Question

A transparent photographic slide is placed in front of a converging lens with a focal length of 2.37 cm. An image of the slide is formed 14.6 cm from the slide.

(a) 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.)

(b) How far is the lens from the slide if the image is virtual? (Enter your answers from smallest to largest starting with the first answer blank. Enter NONE in any remaining answer

blanks.)

Explanation / Answer

a) Use the equation:
1/S1 + 1/S2 = 1/f

The distance between the slide and the real image is:
S1+S2 = 14.6 cm

Plug it into the equation:
1/S1 + 1/S2 = 1/f
1/S1 + 1/(14.6 - S1) = 1/2.37
(14.6 - S1 +S1) / (S1*(14.6 - S1)) = 1/2.37
S1*(14.6 - S1) = 14.6*2.37
14.6*S1-S1² = 34.602
S1² - 14.6*S1 + 34.602 = 0

Quadratic formula:
S1 = 2.97 cm or 11.62 cm

Note both answers for S1 are possible:
If S1 = 2.97cm then S2 = 11.63cm
If S1 = 11.63cm then S2 = 2.97cm

b)  Distances to a virtual image are negative:
i/u + 1/v = 1/f
i/u = 1/f - 1/v = 1/2.37 + 1/14.6 = 0.49
u = 2.04 cm

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