An object is placed 10 m before a convex lens with focal length 5.3 m . Another
ID: 1382318 • Letter: A
Question
An object is placed 10 m before a convex lens with focal length 5.3 m . Another convex lens is placed 22.9 m behind the first lens with a focal length 7 m (see the figure below). Note: Make a ray diagram sketch in order to check your numerical answer.
A. At what distance is the first image from the first lens? Answer in units of m.
B. What is the magnification of the first image?
C. At what distance is the second image from the second lens? Answer in units of m.
D. What is the magnification of the final image, when compared to the initial object?
f2= fi 5.3 m X01 10 m 22.9 m 0 5 10 15 20 25 30 35 40 45 50Explanation / Answer
a. Objectdistance u1 = -10 m ( -ve sign as object is to the left of lens)
imagedistance = v1
focallength f1 = 5.3 m
1/f1 = 1/v1 - 1/u1
1/5.3 = 1/v1 - 1/ ( - 10)
1/v1 = 1/5.3 - 1/10
=> v1 = 5.3* 10 / (10 - 5.3)
= 11.276 m
b. Magnification m1 = -v1 / u1
= -11.276 / ( -10.0)
= 1.128
c. Distance between secondlens and image from firstlens = 22.9 - 11.276
= 11.624 m
distanceof object for secondlens u2 = -11.624 m (-vesign as object is to the left of lens)
1/f2 = 1/v2 - 1/u2
1/7 = 1/v2 - 1/(-11.624)
1/v2 = 1/7 - 1/11.624
v2 = 7* 11.624 / (11.624 - 7)
= 17.596 m
d. Magnification m2 = -v2 / u2
= -(17.596) / ( - 11.624)
= 1.51
totalmagnification m = m1* m2
= 1.128* ( 1.51)
= 0.747
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