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प्रश्न
An object is placed at a distance of 20 cm in front of a concave lens of focal length 20 cm.
- Find the position of the image, and
- the magnification of the image.
उत्तर
- Object distance, u = -20 cm
Focal length, f = -20 cm (concave lens)
lens formula is,
`1/"v" - 1/"u" = 1/"f"`
`therefore 1/"v" = 1/"f" + 1/"u"`
`therefore 1/"v" = 1/-20 + 1/-20 = -2/20`
∴ v = -10 cm
Hence, the image is 10 cm in front of the lens on the same side as the object. - Magnification m = `v/u`
m = `(-10)/-20`
= `1/2`
= 0.5 cm
As m = +ve indicates, the virtual erect image is half the size of the object.
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संबंधित प्रश्न
A spherical mirror produces an image of magnification -1 on a screen placed at a distance of 50 cm from the mirror.
(a) Write the type of mirror.
(b) Find the distance of the image from the object.
(c) What is the focal length of the mirror?
(d) Draw the ray diagram to show the image formation in this case.
When an object is placed at a distance of 50 cm from a concave spherical mirror, the magnification produced is, `-1/2`. Where should the object be placed to get a magnification of, `-1/5`?
In order to obtain a magnification of −2 (minus 2) with a concave mirror, the object should be placed:
(a) between pole and focus
(b) between focus and centre of curvature
(c) at the centre of curvature
(d) beyond the centre of curvature
In order to obtain a magnification of, −0.6 (minus 0.6) with a concave mirror, the object must be placed:
(a) at the focus
(b) between pole and focus
(c) between focus and centre of curvature
(d) beyond the centre of curvature
The lens A produces a magnification of, − 0.6 whereas lens B produces a magnification of + 0.6.
What is the nature of lens B?
Give a scientific reason.
Simple microscope is used for watch repairs.
A lens forms the image of an object placed at a distance 15 cm from it, at a distance 60 cm in front of it. Find the magnification.
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The focal length of a concave lens is 20 cm. At what distance from the lens should a 5 cm tall object be placed so that its image is formed at a distance of 15 cm from the lens? Also calculate the size of the image formed.
What information about the nature of image is erect or inverted, do you get from the sign of magnification + or -?
