An object 2.00 cm high Is placed 38.0 cm to the left of a converging lens having

Question

An object 2.00 cm high Is placed 38.0 cm to the left of a converging lens having a focal length of 28.5 cm. A diverging lens with a focal length of -20.0 cm Is placed 110 cm to the right of the converging lens. (Use the correct sign conventions for the following answers.) (a) Determine the final position and magnitude of the final Image. Image position (b)Is the Image upright or Inverted? (c)Repeat parts (a) and (b) for the case where the second lens Is a converging lens with a focal length of +20.0 cm. Image position

Explanation / Answer

here,

for the first lens

object distance , do1 = 38 cm

focal length , f1 = 28.5 cm

let the image distance be di1

using the lens formula

1/f1 = 1/do1 +1/di1

1/28.5 = 1/38 + 1/di1

di1 = 114 cm

for the seccond lens

the object distance , do2 = 110 - di1 = - 4 cm

focal length , f2 = - 20 cm

let the image distance be di2

using the lens formula

1/f2 = 1/di2 + 1/do2

- 1/20 = 1/di2 - 1/4

di2 = 5 cm

(a)

the final position is 5 cm to the right of diverging lens

m = di1 /do1 * di2 * do2

m = 114/38 * 5/(-4)

m = - 3.75

(b)

the image is inverted

(c)

for the first lens

object distance , do1 = 38 cm

focal length , f1 = 28.5 cm

let the image distance be di1

using the lens formula

1/f1 = 1/do1 +1/di1

1/28.5 = 1/38 + 1/di1

di1 = 114 cm

for the seccond lens

the object distance , do2 = 110 - di1 = - 4 cm

focal length , f2 = 20 cm

let the image distance be di2

using the lens formula

1/f2 = 1/di2 + 1/do2

1/20 = 1/di2 - 1/4

di2 = 3.33 cm

the final position is 3.33 cm to the right of diverging lens

m = di1 /do1 * di2 * do2

m = 114/38 * 3.33/(-4)

m = - 2.5

the image is inverted

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