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प्रश्न
Find the position and magnification of the image of an object placed at distance of 8.0 cm in front of a convex lens of focal length 10.0 cm. Is the image erect or inverted?
उत्तर
Object distance=u=-8cm
Focal length f = 10 cm
Image distance v=?
`1/"v" - 1/"u" = 1/"f"`
`1/"v" - 1/-8 = 1/10`
`1/"v" = 1/10 - 1/8`
`1/"v" = (4 - 5)/40`
`1/"v" = 1/-40`
v = - 40 cm
As the object is placed between the focus and optical center of the lens, the image formed is virtual and erect.
Magnification (m) = `v/u = (-0)/-8`
m = +5.0
The +ve sign shows that the image formed is erect.
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संबंधित प्रश्न
A divergent lens of focal length 30 cm forms the image of an object of size 6 cm on the same side as the object at a distance of 15 cm from its optical centre. Use lens formula to determine the distance of the object from the lens and the size of the image formed.
A student has obtained a point image of a distant object using the given convex lens. To find the focal length of the lens he should measure the distance between the :
(A) lens and the object only
(B) lens and the screen only
(C) object and the image only
(D) lens and the object and also between the object and the image
At what distance from a concave lens of focal length 20 cm, should a 6 cm tall object be placed so that it forms an image at 15 cm from the lens ? Also determine the size of the image formed.
In order to obtain a magnification of, −1.5 with a concave mirror of focal length 16 cm, the object will have to be placed at a distance
(a) between 6 cm and 16 cm
(b) between 32 cm and 16 cm
(c) between 48 cm and 32 cm
(d) beyond 64 cm
Explain what is meant by a virtual, magnified image.
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Give a scientific reason.
Simple microscope is used for watch repairs.
The lens of the eye is flattened when looking at nearby objects.
The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?
What information about the nature of image is erect or inverted, do you get from the sign of magnification + or -?
