Thin lenses. Object O stands on the central axis of a thin symmetric lens. For t
ID: 1440098 • Letter: T
Question
Thin lenses. Object O stands on the central axis of a thin symmetric lens. For this situation, each problem in the table (below) gives object distance p (centimeters), the type of lens (C stands for converging and D for diverging), and then the distance (centimeters, without proper sign) between a focal point and the lens. Find (a) the image distance i and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real (enter 1) or virtual (enter 0), (d) inverted (enter 1) from object O or noninverted (enter 0), and (e) on the same side of the lens as object O (enter 1) or on the opposite side (enter 0).
Thin Lenses. See the setup for these problems. Lens /T1 R/V I/NI Side 50 51 52 53 54 +10 +8.0 19 +12 +12 +45 +25 +22 D, 6.0 D, 12 C, 8.5 C, 16 D, 31 C, 20 C, 35 D, 14 56 57Explanation / Answer
A. Remember This:
For a diverging (D) lens, the focal length value is negative. The object distance p, the image distance i, and the focal length f are related by:
1/f = 1/p + 1/i
The value of i is positive for real images, and negative for virtual images
lateral magnification is m = -i/p
The value of m is positive for upright (not inverted) images, and is negative for inverted images
A.
f = -6, p = 10
using above formula:
i = pf/(p -f) = 10*(-6)/(10 - (-6)) = -3.75
m = -q/p = -(-3.75)/10 = 3.75/10 = 0.375
The fact that the image distance is a negative value means the image is virtual (V).
A positive value of magnification means the image is not inverted (NI).
The image is on the same side as the object
B.
f = -12, p = +8
i = 8*(-12)/(8 + 12) = - 4.8
m = -(-4.8)/8 = +0.6
The fact that the image distance is a negative value means the image is virtual (V).
A positive value of magnification means the image is not inverted (NI).
The image is on the same side as the object.
C.
p = +19, f = 8.5
i = 8.5*19/(19 - 8.5) = +15.38
m = -15.38/19 = -0.809
The fact that the image distance is a positive value means the image is real (R)
A negative value of magnification means the image is inverted (I).
The image is on the opposite side of the object
C.
p = 12, f = 16
i = 12*16(12 - 16) = -48
m = -(-48)/12 = +4
The fact that the image distance is a negative value means the image is virtual (V).
A positive value of magnification means the image is not inverted (NI).
The image is on the same side as the object
you can solve rest of four by same calculations.
comment below if you need further help.
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