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Question
An object 1.5 cm high when placed in front of a concave. the mirror produces a virtual image 3 cm high. If the object is placed at a distance of 6 cm from the pole of the mirror, calculate:
- the position of the image
- the focal length of the mirror.
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Solution
Size of object = h0 = 1.5 cm
Size of image = h1 = + 3 cm [Image formed by a concave mirror is virtual]
Distance of the object from the pole of the mirror = u = - 6 cm
(i) Distance of the image from the pole of mirror = v = ?
Linear magnification = m = `"h"_1/"h"_0 = - "v"/"u"`
`(+3)/1.5 = - "v"/-6`
v = `+ (6 xx 3)/1.5`
v = + 12 cm
So, image is formed at a distance of 12 cm behind the concave mirror.
(ii) Focal length of the concave mirror = f = ?
Using mirror formula:
`1/"u" + 1/"v" = 1/"f"`
`1/-6 + 1/(+12) = 1/"f"`
`1/"f" = - 1/6 + 1/12 = (- 2 +1)/12 = -1/12`
⇒ f = - 12 cm
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