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प्रश्न
A 2.0 cm tall object is placed 40 cm from a diverging lens of focal length 15 cm. Find the position and size of the image.
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उत्तर
A concave lens is also known as a diverging lens.
Focal length of concave lens, f = -15 cm
Object distance from the lens, u = -40 cm
Height of the object, h1 = 2.0 cm
Height of the image, h2 = ?
Using the lens formula, we get:
`1/f=1/v-1/u`
`1/-15=1/v-1/-40`
`1/-15=1/v+1/40`
`1/v=1/-15-1/40`
`1/v=(-8-3)/120`
`1/v=-11/120`
∴ v = - 10.90 cm
Therefore, the image is formed at a distance of 10.90 cm and to the left of the lens.
Magnification of the lens:
Magnification=`"Image distance"/"object distance"="Height of the image"/"Height of the object"`
`v/u=h_2/h_1`
`-10.90/-40=h_2/2`
h_2=+0.54
The height of the image formed is 0.54 cm. Also, the positive sign of the height of the image shows that the image is erect.
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संबंधित प्रश्न
An image that cannot be obtained on a screen is called ______.
Match the items given in Column I with one or more items of Column II.
| Column I | Column II | ||
| (a) | A plane mirror | (i) | Used as a magnifying glass. |
| (b) | A convex mirror | (ii) | Can form image of objects spread over a large area. |
| (c) | A convex lens | (iii) | Used by dentists to see enlarged image of teeth. |
| (d) | A concave mirror | (iv) | The image is always inverted and magnified. |
| (e) | A concave lens | (v) | The image is erect and of the same size as the object. |
| (vi) | The image is erect and smaller in size than the object. |
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