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Light Reflection and Refraction
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13/08/2021 1:43 pm
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A diverging mirror of radius of curvature 40 cm forms an image which is half the height of the object. Find the object and image positions.
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13/08/2021 1:49 pm
R = 40 cm
f = \(\frac{R}{2}\) = \(\frac{40}{2}\) = 20 cm
Image is half the height of the object.
m = -\(\frac{v}{u}\) = \(\frac{h_2}{h_1}\) = \(\frac{1}{2}\)
⇒ u = -2v
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
⇒ \(\frac{1}{v}\) + \(\frac{1}{-2v}\) = \(\frac{1}{20}\)
⇒ \(\frac{1}{v}\) -\(\frac{1}{2v}\) = \(\frac{1}{20}\)
⇒ \(\frac{1}{2v}\) = \(\frac{1}{20}\)
∴ v = 10 cm
u = -2v = -2 x 10 = -20 cm
So, the object is placed 20 cm in front of the mirror and the image is formed 10 cm behind the mirror.