Part A Two converging lenses are placed D = 60.0 cm apart. The initial object is

Question

Part A Two converging lenses are placed D = 60.0 cm apart. The initial object is placed o1 28.7 cm in front of the first lens. The focal length of the second lens is known to be f2 25.3 cm which produces an image on a screen 17.3 cm from the second lens. (Figure 1) What is the focal length of the first lens? cn Submit My Answers Give Up Part B What is the total linear magnification for the system? mtotal 10.058 Submit My Answers Give Up Incorrect Iry Again, 5 attempts remaining Part C If the screen is moved to twice it's distance from the second lens and the lenses are left in place, where should the first object be moved to produce a clear image on the screen? Figure 1 of 1 cn 01 Submit My Answers Give Up Initial Object Part D What is the total linear magnification for this second arrangement? f2 Screen

Explanation / Answer

a)

For the second lens,

f=25.3 cm

v = 17.3 cm

u = ?

using lend formula, 1/f = 1/v + 1/u

we have, 1/25.3 - 1/17.3 = 1/ u

u = -54.71 cm

but D is 60 cm

hence the distance of the image of the first lens will be 60 - 54.71 = 5.29 cm

v = 5.29 cm

u = -28.7 cm

using lens formula for the second lens we have,

1/f = 1/v + 1/u

1/f = 1/5.29 - 1/28.7

f = 6.49 cm

b)

total magnification of the system = m1*m2 = (2.29/28.7)*(17.3/54.71) = 0.025

c) now for the second lens,

f = 25.3 cm

v = 34.6 cm

using lens equation, 1/u = 1/f-1/v = 1/25.3 - 1/34.6

u = 94.13 cm

for the first lens we have,

f=6.49 cm

v = 154.13 cm

using lens equation, 1/u = 1/f - 1/v = 1/6.49 - 1/154.13

u = 6.78 cm

d) magnification: m1*m2 = (154.13/6.78)*(34.6/94.13) = 8.36

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