n the figure, an object of height 2.00 mm is placed 50 cm to the left of a conve

Question

n the figure, an object of height 2.00 mm is placed 50 cm to the left of a converging lens of focal length f_1=+20 cm. A diverging lens of focal length f_2=-10 cm is 15 cm to the right of the first lens.

Part A : Select the location (relative to the second lens) of the final image formed by the pair of lenses.

A) to the left

B) to the right

Part B : Calculate the distance of the final image to the second lens formed by the pair of lenses.

Express your answer in cm using two significant figures.

Part C: Determine nature (real or virtual) of the final image formed by the pair of lenses.

A) real

B) virtual

Part D : Determine orientation (erect or inverted) of the final image formed by the pair of lenses.

A) erect B) inverted

Part E :Calculate the height of the final image formed by the pair of lenses.

Express your answer in mm using two significant figures.

Explanation / Answer

For the converging lens, apply

1/f = 1/p + 1/q

1/20 = 1/50 + 1/q

q = 33.33 cm to the right of the first lens

Since the diverging lens is 15 cm to the right of the converging lens, the image formed from the converging lens is 33.33 - 15 = 18.33 cm behind the lens, so if acts as a virtual image for that lens.

Apply 1/f = 1/p + 1/q

1/-10 = 1/-18.33 + 1/q

q = -22 cm

Since q is negative, that means the final image is virtual and 22 cm to the left of the diverging lens.

So, for part A)

The final image is to the left of the diverging lens (choice A).

For Part B)

The final image is 22 cm to the left of the diverging lens.

For Part C)

Since the final value of q is negative, the final image is virtual.

For Part D)

M for the first lens = -q/p = -33.33/50 = -.667

M for the second lens = -q/p = -(-22)/-18.33) = -1.20

Total magnification = (-.667)(-1.2) = .80

Since the magnification is positive, the final image is erect

For Part E)

M = h'/h

.80 = h'/2

h' = 1.6 mm

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

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