An object is placed (a)6cm (b)20 from a thin convex lens of focal length 10cm.determine the nature,position and magnification of the image in each case.
(a) and (b) are separate problems, not multiple choices.
For the first case, letting di be the image distance from the lens, use the standard lens equation
1/6 + 1/di = 1/f = 1/10
1/di = 6/60 - 10/60 = -4/60 = -1/15
di = -15 cm
The image is "virtual" upright and magnified by a ratio |15/6| = 2.5
Now you do case b
i dn't no the answer for case b pls help me
To determine the nature, position, and magnification of the image formed by a thin convex lens, we can use the lens formula and the magnification formula.
The lens formula is given by:
1/f = 1/v - 1/u
Where:
f is the focal length of the lens,
v is the image distance from the lens (positive if the image is on the opposite side of the lens from the object),
u is the object distance from the lens (positive if the object is on the same side as the incident light).
The magnification formula is given by:
m = -v/u
Where:
m is the magnification.
Now let's solve the problem:
(a) When the object is placed 6 cm from the lens:
Given:
f = 10 cm
u = -6 cm (negative because the object is on the same side as the incident light)
Substituting these values into the lens formula:
1/10 = 1/v - 1/-6
Multiplying through by 10v(-6) to get rid of the fractions:
-v(-6) = -6v = -60v
60v + 6v = -60
66v = -60
v = -60/66
v ≈ -0.91 cm
Now, calculate the magnification using the magnification formula:
m = -v/u
m = -(-0.91) / -6
m ≈ 0.91 / 6
m ≈ 0.15
Therefore, in this case, the image formed by the lens is virtual, located approximately 0.91 cm on the same side as the object, and has a magnification of approximately 0.15.
(b) When the object is placed 20 cm from the lens:
Given:
f = 10 cm
u = -20 cm (negative because the object is on the same side as the incident light)
Substituting these values into the lens formula:
1/10 = 1/v - 1/-20
Multiplying through by 10v(-20) to get rid of the fractions:
-v(-20) = -20v = -400v
400v - 20v = -20
380v = -20
v = -20/380
v ≈ -0.053 cm
Calculate the magnification using the magnification formula:
m = -v/u
m = -(-0.053) / -20
m ≈ 0.053 / 20
m ≈ 0.00265
Therefore, in this case, the image formed by the lens is virtual, located approximately 0.053 cm on the same side as the object, and has a magnification of approximately 0.00265.