A lens is placed 14 cm from a 4 cm high object, and the image is observed 22 cm

Question

A lens is placed 14 cm from a 4 cm high object, and the image is observed 22 cm from the lens. a. What is the object distance? b. What is the image distance? c. What is the focal length of the lens? (Use the lens equation) d. What is the expected magnification from the lens? (Use the magnification equation) e. What is the height of the image? f. Is the image upright or inverted? Leaving the object and first lens in placed, a second lens, with focal length 5 cm, is placed 25 cm from the first lens. a. What is object distance for the second lens? b. What is the image distance from the second lens? c. What is the magnification from (just the) second lens? d. What is the total magnification of the pair of lenses? e. What is the final image height (from both lenses)? f. Is the final image upright or inverted? A lens with focal length 10 cm is placed 5 cm from an object. a. What is the image distance? b. Is this image real or virtual? c. If a second lens, of focal length 5 cm, is placed 10 cm from the first lens, is the image viewed through both lenses real or virtual?

Explanation / Answer

1. a Object distance u = 14 cm

b) image distance v =22 cm

c) focal length f = uv /( u+v) = 14 x 22 / (14+22) = 8.56cm

d) Magnification m1 = - v/u = - 22/14 = - 1.57

e) Magnification m1 = - v/u = hi/ho

height of the image hi = mi x ho = -1.57 x 4 = - 6.28 cm

f) As the magnification is negative it is an inverted image

2. a) Obect distance u = 3cm

b) u= 3cm focal length f = 5cm

image distance v = uf/ u-f = 3x5 / 3-5 = - 15/2 = - 7.5cm

c) magnification m2 = - v/u = 7.5/3 = 2.5cm

d) Total magnification M = m1 x m2  = - 6.28 x 2.5 = - 15.7

e) Magnification M = hi/ho  

height of the image hi = M x ho = - 15.7 x 4 = - 62.8 cm

f) since the magnification is negative it is a inverted image

3. a) given f = 10cm u = 20cm

image distance v =uf/u-f = 10 x 5 / 5-10 = - 10cm

b) It is a virtual image since v is negative

c) now f = 5cm u= 20cm

image distance v = uf/u-f = 5 x 20/ 20-5 = 6.67 cm

It is a real image

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