A 3.5-cm tall object is placed 19.4 cm in front of a converginglens. A real imag

Question

A 3.5-cm tall object is placed 19.4 cm in front of a converginglens. A real image is formed 55.9 cm from the lens.

(a) What is the focal length of the lens?

1 cm

(b) If the original lens is replaced with a lens having a focallength of 28.8, what are the image position, size and orientation?Enter negative answers where appropriate.

Image distance = 2cm
Image Height = 3 cm
Orientation: 4--Pull-Down --uprightinverted (1 allowed attempt)

Explanation / Answer

(a) 1 / f = 1 / do + 1 /di      ==> f = do*di / [ do + di] Where do = 19.4 cm             di = - 55.9 cm ==> f = 19.4 * (-55.9) / [ 19.4 - 55.9]           = 29.71 cm (b) if f = 28.8 cm, do = 19.4 cm            imageposition di = do * f / [do - f]                                      = 19.4 * 28.8 / [19.4-28.8]                                      = -59.43 cm      size hi = - ho*di / do               = - 3.5 * -59.43 / 19.4               = 10.72 cm

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