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? == 6.74E+01 cm and image = real, inverted, smaller, and behind lens

HELP --

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

Is the new 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.)

Explanation / Answer

Solution:

Answer to the first part, that is 6.74E+1 cm is correct.

Now the object is moved to 27.5 cm in front of the lens, thus,

object distance p = 27.5 cm

focal point , f = 39.5 cm

Using the thin lens formula,

1/f = 1/p + 1/i

1/i = 1/f – 1/p

1/i = 1/(39.5cm) – 1/(27.5cm)

1/i = (27.5 – 39.5) /1086.25

1/i = -12/1086.25

i = - 1086.25/12 cm

i = - 90.52 cm

Thus the image distance with negative sign indicates that the image is in front of the lens (that is the same side of lens as that of the object) and is thus virtual. (Real images form only behind the lens with positive value of i)

magnification m = -i/p = - (-90.52cm)/(27.5cm) = 3.29

Since the magnification is a positive number, it is an upright image.

Since the magnification is greater than 1, the image is larger than the object.

Thus the properties of the image are: virtual, upright, larger and front.

And the answer is; VULF

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