An object 1 cm high produces a real image 1.5 cm high, when placed at a distance of 15 cm from a concave mirror. Calculate : (i) the position of the image, (ii) focal length of the concave mirror.
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Answered by
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Formula = h' u / h = v
- v = (1.5) (-15) / 1
- v = - 22.5
v = 22.5
The image is positive, therefore it is a virtual and erect image.
∴ the image is formed at a distance of 22.5 cm behind the mirror.
1 / v + 1 / u = 1 / f
1 / f = 1 / 22.5 + 1 / (- 15)
1 / f = 1 / 22.5 -1 / 15
1 /f = (15 - 22.5) / 337.5
1 / f = -7.5 / 337.5
f = 337.5 / - 7.5
f = - 45 cm
∴ the focal length of the mirror is 45 cm.
- v = (1.5) (-15) / 1
- v = - 22.5
v = 22.5
The image is positive, therefore it is a virtual and erect image.
∴ the image is formed at a distance of 22.5 cm behind the mirror.
1 / v + 1 / u = 1 / f
1 / f = 1 / 22.5 + 1 / (- 15)
1 / f = 1 / 22.5 -1 / 15
1 /f = (15 - 22.5) / 337.5
1 / f = -7.5 / 337.5
f = 337.5 / - 7.5
f = - 45 cm
∴ the focal length of the mirror is 45 cm.
Answered by
1
Now we have to find focal length.
1/v+1/u=1/f
1/22.5+1/(-15)=1/f
15-22.5/337.5=1/f
-7.5/337.5=1/f
f=337.5/-7.5
f=-45 cm
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