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Question
An object of height 4.25 mm is placed at a distance of 10 cm from a convex lens of power +5D. Find (i) the focal length of the lens, and (ii) the size of the image.
Numerical
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Solution
Object height, h = 4.25 mm = 0.425 cm (1 cm = 10 mm)
Object distance, u =-10 cm
Power, P = +5 D
Focal length, f = ?
Image distance, v = ?
Image height, h' = ?
Power, P `=1/f`
`f=1/p=1/5=0.2m =20cm `
Using the lens formula, we get:
`1/v-1/u=1/f`
`1/v-1/-10=1/20`
`1/v+1/10=1/20`
`1/v=1/20-1/10`
`1/v=-1/20`
∴ v=-20cm
Now, magnification, m = `(h')/h=v/u`
Substituting the values in the above equation, we get:
`(h')/0.425=(-20)/-10`
`h'=2X0.425=0.85cm=8.5mm`
Thus, the image is 8.5 mm long; it is also erect and virtual.
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