A candle 7.90 cm high is placed in front of a thin converging lens of focal leng

Question

A candle 7.90 cm high is placed in front of a thin converging lens of focal length 39.5 cm.What is the image distance i when the object is placed 95.5 cm in front of this lens?

Is the image real(R) or virtual(V); upright(U) or inverted(I); larger(L) or smaller(S) or unchanged(UC); in front of the lens(F) or behind the lens(B)? Answer these questions in the order that they are posed. For example, if the image is real, inverted, larger and behind the lens then enter `RILB'. (Please note that the OBJECT is defined to be in FRONT of the lens in this problem.)

The object is now moved to 27.5 cm in front of the lens, what is the new image distance i?

Explanation / Answer

1/v - 1/u = 1/ f

1/v = 1/ f + 1/u

= 1/ 39.5 + 1/ -95.5

v = 67.36160

m = v /u =  67.36160 / -95.5 = -0.70 minus sign means inverted

height of image = Ho * m = 7.90*0.705 = 5.57 cm

its a real image , inverted ; smaller ;behind the lense

now if it is moved to 27.5 then

u = 95.5-27.5 = 68 cm

1/v - 1/u = 1/ f

1/v = 1/ f + 1/u

= 1/ 39.5 + 1/ -68

v = 94 .24 cm

m = v/u = 94 .24 / 95.5 = - 0.98 .........inverted

height of image = 0.98*7.9 = 7.79 cm

real : inverted : smaller : behind the lens

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