Problem 9: Consider the compound optical system shown in the diagram, where two

Question

Problem 9: Consider the compound optical system shown in the diagram, where two thin lenses of focal lengths 5.5 cm (left lens) and 32 cm (right lens) are separated by a distance 27 cm.

Part (1) Does the image formed by the first lens serve as a real or a virtual object for the second lens?

MultipleChoice : 1) Virtual 2) Real

Part (2) What is the image distance, in centimeters, for the second lens?

Numeric : A numeric value is expected and not an expression. di2 = __________________________________________

Part (3) What is the magnification of the second lens?

Numeric : A numeric value is expected and not an expression. M2 = __________________________________________

Part (4) What is the total magnification of this compound optical system?

Numeric : A numeric value is expected and not an expression. M = __________________________________________

Part (5) Is the image created by the second lens real or virtual? Is it upright or inverted?

MultipleChoice : 1) Real and inverted. 2) Real and upright. 3) Virtual and inverted. 4) Virtual and upright..

Explanation / Answer

Given,

f1 = 5.5 cm and f2 = 32 cm ;

s = 27 cm.

From figure do = s = 27 cm.

1)Virtual object for the second lens.

2)we know from the lens equation that,

1/f = 1/i + 1/o

i = o x f / ( o - f)

di1 = do x f1 / (do - f1) = 27 x 5.5 / ( 27 - 5.5) = 6.91 cm

do2 = s - di1 = 27 - 6.91 = 20.1 cm

Again on using lens equation, di2 = do2 x f2 / (do2 - f2) = 20.1 x 32 / ( 20.1 - 32) = -54.1 cm

3)m1 = -di1/do1 ; m2 = -di2/do2

m1 = -6.91/27 = -0.26

m2 = -(-54.1)/20.1 = +2.7

4)total magnification will be:

M = m1 x m2 = -0.26 x 2.7 = -0.702

5)If the image distance di2 is positive then its real else virtual. If the overall magnification is positive the image is upright.

Since di2 = -54 cm, image is virtual and since M = -0.702 image is inverted.

3)Virtual and inverted is correct choice.

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