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प्रश्न
The power of a lens is +4D. Find the focal length of this lens. An object is placed at a distance of 50 cm from the optical centre of this lens. State the nature and magnification of the image formed by the lens and also draw a ray diagram to justify your answer.
उत्तर
Focal length = `1/"P"= 1/4` = 0.25 m or 25 cm. It is a convex lens as power and focal length are positive.
Focal length, f = +25 cm
Object distance, u = -50 cm
Image distance, v = ?
Image height, I = ?
Using lens formula,
`1/"f"=1/"v"-1/"u"`
`1/25=1/"v"-1/((-50))`
`1/25 + 1/((-50)) =1/"v"`
`(2 - 1)/50 =1/"v"`
`1/"v" = 1/50`
v = +50 cm
Also, magnification, m = `"v"/"u"`
Hence, m = `((50))/-(50)= -1`
Image formation occurs thus 50 cm in front of the lens. Due to a magnification of 1, the image is the same size as the object. The image is inverted as evidenced by the negative magnification.
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