The magnification of an image formed by a lens is –1. If the distance between the object and its image is 60cm, what is the distance of the object from the optical centre of the lens? Find the nature and focal length of the lens. If the object is displaced 20cm towards the optical centre of the lens, where would the image be formed and what would be its nature? Draw a ray diagram to justify your answer.
CBSE FOREIGN - SET 3 2017
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m = -1, Hence the image is real and the lens convex. As m = -1, u = v u + v = 60 cm (given) i.e. = 4f = 60 cm. When object is at 2F, image is also at 2F distance i.e. f = +15 cm On displacing the object by 20 cm towards the lens u = -10 cm$\text{As}\text{ }{\frac{1}{f}}=\frac{1}{v}-{\frac{1}{u}},{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }{\frac{1}{f}}=\frac{1}{v}+{\frac{1}{u}}}$
$\text{Or}\text{ } \frac{1}{v}=\frac{1}{+15{\text{cm}}}+\frac{1}{-10{\text{cm}}}=\frac{-1}{30{\text{cm}}}$
$\text{Or}\text{ }{v}=-30{\text{cm}}$
Nature of the image will be virtual.
$\text{Or}\text{ } \frac{1}{v}=\frac{1}{+15{\text{cm}}}+\frac{1}{-10{\text{cm}}}=\frac{-1}{30{\text{cm}}}$
$\text{Or}\text{ }{v}=-30{\text{cm}}$
Nature of the image will be virtual.
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