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An object 1.0 cm in height, is placed at a distance of 25.0 cm in front of a concave mirror of focal length 20.0 cm, on its principal axis. Following New Cartesian Sign Convention, the image is formed at v = ______ and its height hI = ______.
The correct answer is -100 cm; -4.0 cm.Concept:Mirror formula:\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\), where f = focal length, v = image distance and u = object distance.Magnification: m = \(\frac{Hi}{Ho} = \frac{-v}{u}\) , where Hi = height of image, Ho = height of object, v = image distance, and u = object distance.Concave mirror: It is part of a whole spherical mirror that reflects light from the hollow part.It is a converging mirror because it converges the incident rays at a point.Explanation:Given, u = -25 cm f = -20 cm h = 1 cm \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)\(\frac{1}{-20} = \frac{1}{v} + \frac{1}{-25}\)\(\frac{1}{-20} = \frac{1}{v} - \frac{1}{25}\)\(\frac{1}{v} = \frac{1}{-20} + \frac{1}{25} \)\(\frac{1}{v} = \frac{-5 + 4}{100} \)\(\frac{1}{v} = \frac{-1}{100} \)v = -100 cmm = \(\frac{Hi}{Ho} = \frac{-v}{u}\) = \(\frac{Hi}{1} = \frac{-(-100)}{-25}\) = Hi = - 4 cmFollowing New Cartesian Sign Convention, the image is formed at v = -100 cm and its height hI = -4 cm.
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