Question:
In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for
which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.
Solution:
u = -x1
v = +(D – x1)
1/D – x1 – 1/(-x1) = 1/f
u = -x2
v = +(D – x2)
1/D – x2 – 1/(-x2) = 1/f
D = x1 + x2
d = x2 – x1
x1 = D – d/2
D – x1 = D + d/2
u = D/2 + d/2
v = D/2 – d/2
m1 = D – d/D + d
m2/m1 = (D+d/D-d)2
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.