EWhen two lenses are used in combination, the first one forms an image that then

Question

EWhen two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.30 cm -tall object is 57.0 cm to the left of a converging lens of focal length 40.0 cm . A second converging lens, this one having a focal length of 60.0 cm , is located 300 cm to the right of the first lens along the same optic axis.

A)Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm .

B) I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

Explanation / Answer

Lens 1

u = 57 cm

f = 40 cm

A)

v = fu/ (u - f) = 40*57 / (57-40) = 134.12 cm

So image is 134.12 cm right of lens 1

Height of image = - object height * v / u = -1.3*134.12/57 = - 3.0588 cm = 3.0588 cm inverted

B)

Image distance with respect to lens 2 = 300 - 134.12 = 165.88 cm to the left of lens 2

focal length = 60 cm

v = fu/ (u - f) = 60*165.88 / (165.88-60) = 94 cm

So image is 94 cm right of lens 2

Height of image = - object height * v / u = 3.0588*94/165.88 = 1.733 cm = 1.733 cm erect

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

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