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An object is placed at a distance of 20.0 cm from a concave mirror of focal length 15.0 cm. (a) What distance from the mirror a screen should be placed to get a sharp image? (b) What is the nature of the image? |
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Answer» Solution :Give: f = - 15 cm, `mu = 20 cm` (a) Mirror equations, `(1)/(v) + (1)/(u) = (1)/(f)` Rewriting to find `v, (1)/(v) = (1)/(f) - (1)/(u)` Substituting for f and u, `(1)/(v) = (1)/(-15) - (1)/(-20)` `(1)/(v)=((-20)-(-15))/(300)xx(-5)/(300)=(-1)/(60)` v = - 60.0 cm As the image is formed at 60.0 cm to the left of the concave mirror, the screen is to be placed at distance 60.0 cm to the left of te concave mirror. (b) MAGNIFICATION, m = `(h.)/(h) = - (v)/(u)` `m = (h.)/(h) = ((-60))/((-20)) = - 3` As the sign of magnification is NEGATIVE, the image is inverted. As the magnitude of magnification is 3, the image is enlarged three times. As the image is formed to theleft of the concave mirror, the image is real. |
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