how far from a convex lens of focal length 20 cm would you place an object toget an image enlarged 3 times
Answers
Answer:
u= -2 x 20/3 = -40/3 cm. Answer: Between the centre of the curvature and its focus, if the image is real
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Question: How far from a convex lens of focal length 20cm would you place an object to get an image enlarged 3 times?
Answer:
There could be two position of object
There could be two position of object i) For real image Given , M = -3
There could be two position of object i) For real image Given , M = -3 m = v/u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual image
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3m= v/u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3m= v/u3= v/u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3m= v/u3= v/uv=3u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3m= v/u3= v/uv=3uby lens formula
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3m= v/u3= v/uv=3uby lens formula1/f = 1/v - 1/u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3m= v/u3= v/uv=3uby lens formula1/f = 1/v - 1/u1/20 = 1/3u - 1/u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3m= v/u3= v/uv=3uby lens formula1/f = 1/v - 1/u1/20 = 1/3u - 1/u1/20= -2/3u
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3m= v/u3= v/uv=3uby lens formula1/f = 1/v - 1/u1/20 = 1/3u - 1/u1/20= -2/3uu= -2 x 20/3 = -40/3 cm.
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3m= v/u3= v/uv=3uby lens formula1/f = 1/v - 1/u1/20 = 1/3u - 1/u1/20= -2/3uu= -2 x 20/3 = -40/3 cm.therefore the two positions of the object are:
There could be two position of object i) For real image Given , M = -3 m = v/u-3 = v/uv = -3u ...................... (i) by lens formula 1/f = 1/v - 1/u 1/20 = -1/3u - 1/u1/20 = -4/3uu= -4 x 20/3 = - 80/3 cm for virtual imagem= +3m= v/u3= v/uv=3uby lens formula1/f = 1/v - 1/u1/20 = 1/3u - 1/u1/20= -2/3uu= -2 x 20/3 = -40/3 cm.therefore the two positions of the object are:-80/3 cm and -40/3 cm.
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