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) A candle of 5 cm in length is held 15 cm away from converging mirror of focal length [3]10 cm. Find the position, size and the nature of the image formed. Draw the ray diagram. |
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Answer» Answer: Given that The height of object = 5cm POSITION of object, u = – 25cm The focal length of the lens, f = 10 cm We need to find The position of the IMAGE, v =? Size of the image Nature of the image Formula We know that 1/v – 1/u = 1/f Substituting the known values in the above equation we get, 1/v + 1/25 = 1/10 => 1/v = 1/10 – 1/25 => 1/v = (5 – 2)/50 Hence, 1/v = 3/50 So, v= 50/3 = 16.66 cm Therefore, the distance of the image is 16.66 cm on the OPPOSITE side of the lens. Now, we know that Magnification = v/u Hence, m = 16.66/-25 = -0.66 Also, we know that m= height of image/height of the object Or, -0.66 = height of image / 5 cm Hence, height of image = -3.3 cm The negative sign of the height of the image DEPICTS that an inverted image is formed. So, the position of image = At 16.66 cm on the opposite side of the lens Size of image = – 3.3 cm at the opposite side of the lens Nature of image – Real and inverted |
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