An object is placed 10 m before a convex lens with focal length 5.7 m . Another

Question

An object is placed 10 m before a convex lens with focal length 5.7 m . Another concave lens is placed 9.88 m behind the first lens with a focal length ?1.8 m (see the figure below). Note: Make a ray diagram sketch in order to check your numerical answer.

A. At what distance is the first image from the first lens? Answer in units of m.

B. What is the magnification of the first image?

C. At what distance is the second image from the second lens? Answer in units of m.

D. What is the magnification of the final image, when compared to the initial object?

Explanation / Answer

Given that,

o1 = 10 m ; f1 = 5.7 m

D = 9.88 m and f2 = - 1.8 m

A) we know that,

1/f = 1/i + 1/o

i = o x f / (o - f)

i1 = o1 x f1 / (o1 - f1) = 10 x 5.7 / (10 - 5.7) = 57/4.3 = 13.26

hence the distance of first image = i1 = 13.26 meters to the right of first lens.

B)Magnitfication is given by:

M = -i / o

M1= -i1/o1 = -13.26/10 = -1.326

Hence M1 = -1.326

C)we already got i1 = 13.26 m now, this will act as the object for the second lens, so we have

o2 = D - i1 = 9.88 - 13.26 = -3.38 m

again we will use the same expression

I2 = o2 x f2 / (o2 - f2) = (-3.38) (-1.8) / (-3.38 - (-1.8) = 6.084/1.58 = - 3.85m

Hence i2 = -3.85 m to the left of second lens (bwteen the two lenses)

D)M2 = -i2 /o2 = - (-3.85) / (-3.38) = - 1.14

Hence M2 = -1.14

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