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

Question

Visitors at a science museum are invited to sit in a chair to the right of a full-length diverging lens (f1 = -2.80 m) and observe a friend sitting in a second chair, 2.00 m to the left of the lens. The visitor then presses a button and a converging lens (f2 = +3.70 m) rises from the floor to a position 1.70 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.

Explanation / Answer

A:
We have been given that:
Focal length of the diverging lens f1 = - 2.80 m
Focal length of the converging lens f2 = 3.7 m
object distance for the diverging lens o1 = 2 m
And,
distance between the two lenses L = 1.70 m
let,i1 be the image distnce for diverging lens.
From lens formula
1/f1 = 1/o1 + 1/i1
Hense,
image distance,i1 = -1.167 m
So,magnification of the diverging lens :
M1 = - i1/o1= -(-1.167) / 2 = 0.584
Now,
B:
object distance for converging lens is :
o2 = L - i1 = 1.7 +1.167 = 2.867 m
from lens formula
1/f2 = 1/o2 + 1/i2
Hense,
image distance,i2 = (3.7)(2.867) / (2.867-3.7) = -12.74 cm
magnification of the converging lens is
M2 = - (-12.74) /2.867 = 4.44
Hence ,
overall magnification of the system of lenses is:
M = M1 M2 = 0.584*4.44 = 2.593

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.