Snow Intro/Instruuctions An object of height 2.9 cm is placed at 23 cm in front

Question

Snow Intro/Instruuctions An object of height 2.9 cm is placed at 23 cm in front of a diverging lens of focal length. f18em. Behind the diverging lens, there is a converging lens of focal length, f - 18cm. The distance between the lenses is 3 cm. In the next few steps, you will find the location and size of the final image. Hint: Use + = and m =--to work through each lens stage one at a time. The location of the di do f image from the first lens becomes the location of the object for the second lens (but image distance from the first lens is not necessarily equal to the object distance to the second lens) a. Where is the intermediate image formed by the first diverging lens? Image distance from first lens is cm. (Use the sign to indicate which side the image is on; positive sign il means image is on the side of outgoing rays, and negative sign means image is on the side opposite to the outgoing rays.) dy2 hy b. Where is the final image formed by the second converging lens? Image distance from second lens is cm. (Use the sign to indicate which side the image is on) c. How large is the intermediate image formed by the first diverging lens? Intermediate image height is cm. (Use the sign to indicate whether the image is upright (positive) or inverted (negative).) d. How large is the final image formed by the second converging lens? Final nnage height is ha = em. (Use the sign to indicate whether the image is upright or inverted)

Explanation / Answer

Ans:-

Given data h = 2.9cm ,d0 = 23cm,f= -18cm

A] 1/di + 1/d0 = 1/f

1/di + 1/23 = -1/18

1/di = -41/414

di1=- 414/41 =-10.1cm

b] here d0 = -13.1cm f= 18cm

d0 = di1-3cm = -10.1 -3 = -13.1cm

1/di + 1/d0 = 1/f

1/di -1/13.1=1/18

1/di = 1/18 + 1/13.1 = 31.1/235.8

di2 = 235.8/31.1 =7.58cm

c] m = -di1/d0 =-(-13.1/23) =0.57

|m| = h’/h

h= 2.9cm

0.57 = h’ /2.9

hi1’ =1.65cm

d] m = -di2/d0 = - 7.58/-13.1 = 0.58

|m| = h’i2 / h = h’i2/1.65

0.58*1.65 = h’i2= 0.96cm

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