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 2.53 cm. An image of the slide is formed 12.2 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 you answers from smallest to largest starting with the first answer blank. Enter NONE in any remaining answer blanks.)

Explanation / Answer

Focal length of converging lens, f = +2.53 cm.

The lens forms an image of the slide 12.2 cm from the slide.

a) If the image formed is real, then let the object distance be 'u' and the image distance:

v = u + 12.2 cm

Now, using the lens formula:

(1/v) - (1/u) = (1/f)

We get:

(1/(u + 12.2)) - (1/u) = (1/2.53),

This gives a quadratic equation:

u2 + 12.2u + 30.86 = 0.

The solution of above quadratic equation being, u = -3.58 or -8.61

Thus, there are two possibilities for object distance:

nearer possible distance = 3.58 cm

farther possible distance = 8.61 cm

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