Visitors at a science museum are invited to sit in a chair to the right of a ful

ID: 1294938 • Letter: V

Question

Visitors at a science museum are invited to sit in a chair to the right of a full-length diverging lens (f1 = -3.10 m) and observe a friend sitting in a second chair, 1.90 m to the left of the lens. The visitor then presses a button and a converging lens (f2 = +4.30 m) rises from the floor to a position 2.40 m to the right of the diverging lens, allowing the visitor to view the friend through both lenses at once. Find (a) the magnification of the friend when viewed through the diverging lens only and (b) the overall magnification of the friend when viewed through both lenses. Be sure to include the algebraic sign (+ or -) with your answers.

I have already firgured out part (a) and found a magnification through the diverging lens only as 0.62 and it is correct. I can not figure out part (b) overall magnification.

Explanation / Answer

Please Solve it similarly like this

QUESTION:

Visitors at a science museum are invited to sit in a chair to the right of a full-length diverging lens (f1 = -3.06 m) and observe a friend sitting in a second chair, 1.88 m to the left of the lens. The visitor then presses a button and a converging lens (f2 = 4.08 m) rises from the floor to a position 1.75 m to the right of the diverging lens, allowing the visitor to view the friend through both lenses at once.

(a) Find the magnification of the friend when viewed through the diverging lens only.
(b) Find the overall magnification of the friend when viewed through both lenses.

ANSWER:

To find the image of the single lens use

1/f1 = 1/s1 + 1/s'1

or 1/s'1 = 1/f1 - 1/s1

or s'1 = f1*s1/(s1 - f1) = (-3.06*1.88)/(1.88 - -3.06) = -1.16m (The negative sign means the image is to the left of the lens and is virtual)

so m1 = -s'1/s1 = --1.16/1.88 = 0.619

b) Now the image of the first lens becomes the object of the second so s2 = 1.75 + 1.16 = 2.915m

so 1/f2 = 1/s2 + 1/s'2

or 1/s'2 = 1/f2 - 1/s2

or s2' = f2*s2/(s2 - f2) = (4.08*2.915)/(2.915 -4.08) = -10.2m

so the magnification of the second lens m2 = -s2/s'2 = --10.2/2.915 = 3.50

so the overall magnification = m1*m2 = 0.619*3.50 = 2.17

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Visitors at a science museum are invited to sit in a chair to the right of a ful

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Visitors at a science museum are invited to sit in a chair to the right of a ful

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