The figure below shows an object placed a distance d_o1 from one of two convergi

Question

The figure below shows an object placed a distance d_o1 from one of two converging lenses separated by s = 1.00 m. The first lens has focal length f_1 = 20.0 cm, and the second lens has focal length f_2 = 42.0 cm. An image is formed by light passing through both lenses at a distance d_i2 = 14.0 cm to the left of the second lens. (Include the sign of the value in your answers.) What is the value of d01 that will result in this image position? Cm Is the final image formed by the two lenses real or virtual? real virtual What is the magnification of the final image? Is the final image upright or inverted? upright inverted

Explanation / Answer

a) At the first lens

1 / do1 + 1 / di1 = 1 / f1 --->   do1 = ( di1 f1 ) / ( di1 - f1 ) (1)

the image of the first lens is the object at the second lens, and its position relative to the second lens is

do2 = S - di1 (2)

at the second lens

1 / do2 + 1 / di2 = 1 / f2 --->   do2 = ( di2 f2 ) / ( di2 - f2 ) (3)

the image of the second lens is virtual then di2 = -14.0 cm

do2 = ( (-14.0 cm) * 42.0 cm ) / ( -14.0 cm - 42.0 cm )

do2 = 10.5 cm

from equation (2)

di1 = S - do2

di1 = 100 cm - 10.5 cm

di1 = 89.5 cm

hence, from equation (1)

  do1 = ( 89.5 cm * 20.0 cm ) / ( 89.5 cm - 20.0 cm )

  do1 = 25.8 cm

b) the final image is virtual because is to the left of the second lens

c) the magnification is

M = M1 M2

M = ( - di1 / do1 )( - di2 / do2 )

M = ( di1 di2 )/ ( do1 do2 )

M = ( 89.5 cm * (-14.0 cm) )/ ( 25.8 cm * 10.5 cm )

M = -4.63

d) since the magnification is negative, the image is inverted

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