A lens is creating a virtual image of a candle. which is placed at the distance

Question

A lens is creating a virtual image of a candle. which is placed at the distance of 10 0 cm from the lens The image is formed at 3 6 cm from the lens. Is it diverging or converging lens? What is its focal length? Draw the ray diagram below For the full credit show the focal points. draw at least two rays. and the image Is it upright or inverted? Next the candle is moved half as far away to the lens. Does the image move closer or farther away from the km? By how many centimeters does the position of the image change? Does the candle appear smaller or larger than before? How much smaller or larger does K look now? Specifically what is the ratio of the new magnification to the old one?

Explanation / Answer

Part 1. using formula we can find focal length

1/v + 1/u =1/f

1/0.036 + 1/0.1=1/f , f=0.026m= 2.6cm, focal length is positive so it is convex lens,converging lens.

part c. image move far away from lens .and using lens formula we can calculate new position of image.

u= object distance =0.05m , f=0.026m , v=image position

put these values in above lens eqution to calculate final image position.

change in position =final posion - 3.6cm.

Part d. image will be larger larger, because image distance increases. this can be seen by following formula

magnification =-v/u

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