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(a) With the help of a ray diagram explain why a concave lens diverges the rays of a parallel beam of light. (b) A 2.0 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15 cm. At what distance from the le...

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(a) With the help of a ray diagram explain why a concave lens diverges the rays of a parallel beam of light.

(b) A 2.0 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15 cm. At what distance from the lens, should the object be placed so that it forms an image 10 cm from the lens? Also find the nature and the size of image formed.

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(a) When a parallel beam of light incident on a front face of concave lens, each ray of light will refract towards the normal to the surface as it moves from rarer to denser medium and travels in a straight line inside the lens until it reaches the back face of the lens. At the back face boundary, each ray of light will again refract and bend away from the normal to the surface as it moves from denser to rarer medium. The course of ray of light is shown in the following figure.

Thus, because of the concave shape of both the faces, the double concave lens diverge the rays of parallel beam of incident light.
(b) ho = 2.0 cm, f = -15 cm
v = -10 cm as concave lens always forms virtual image.
Using, 1/f = 1/v - 1/u, we get
1/u = 1/v - 1/f = 1/-10 - 1/-15
= -15 + 10/150
u = -150/5 = -30 cm
Using, m = h1/ho = v/u, we get
m = -10/-30 = 1/3 < 1, we get
h1 = 2 x -10/-30 = 2/3
So, image is virtual, erect and diminished.
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