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Can you please please answer part (c &d) of this problem. Question 21: Image For

ID: 1585931 • Letter: C

Question

Can you please please answer part (c &d) of this problem. Question 21: Image Formation By Multiple Lenses (20 pts) Consider an arrangement of a converging lens of focal length fi and a diverging lens of focal length-fa U is defined to be a positive number; necessary signs are indicated directly on expressions). The converging lens is "stronger" than the diverging lens, that is, f, f + h). The coordinate axis is arranged so that the converging lens is placed at the location x = 0, and the diverging lens is placed at the location x +D. a) If an object is placed at the location x 3fi and viewed through the diverging lens (and through the converging lens, visible within the diverging lens), find where the final image appears to be. Give the x position of the final image. Show your work, clearly explaining each step (as usual, you are graded on the quality of both your answer and the work shown). Lmagt formid by -12+242-34 20-34. b) What is the linear magnification of this arrangement? Give your answer in terms of given quantities fi. fa, and/or D. Show your work, clearly explaining each step. :/ /uD-3f.):W

Explanation / Answer

c. when the smoky screen is placed on the location of image made by the converging lesn, since the converging lens makes a real image, we will get a sharp focused image on the screen, but since it is diffusive transparent, the light passing through it will diffuse and hence ht einfal image made by the divwrging lens will be diffused

when the screen is placed at the location of the image formed by the diverging lens, we will not see any omage on tghe screen since the diverging lens forma a virtual image which can be seen but cannot be obtained on any screen

d. as D decreases from +D to 0,

from the formula for v2 we see

V = (2D - 3f1)2f1/(2D + f1)

for f2 = 2f1

hence

for D = 0

V = -6f1 ( hence the final image is virtual)

hence

for all the decreasing vakue of D, the value of V remains -ve and keeps on decreasing