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An object is placed at a distance of 40 40 40 cm from a concave mirror of focal length 15 15 15 cm. If the object is displaced through a distance of 20 20 20 cm towards the mirror, the displacement of the image will be
medium
Ray Optics and Optical Instruments
2018
physics
30 30 30 cm towards the mirror
36 36 36 cm away from the mirror
30 30 30 cm away from the mirror
36 36 36 cm towards the mirror
Explanation To solve this problem, we will use the mirror formula: 1 f = 1 v + 1 u \\
\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\
f 1 = v 1 + u 1
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where
is the focal length,
is the image distance, and
is the object distance.
Initially, the object is placed at
cm (negative sign because the object is in front of the mirror).
The focal length of the concave mirror is
cm.
Using the mirror formula:
1 − 15 = 1 v 1 + 1 − 40 \\
\frac{1}{-15} = \frac{1}{v_1} + \frac{1}{-40} \\
− 15 1 = v 1 1 + − 40 1 Rearrange to solve for
v 1 : 1 v 1 = 1 − 15 + 1 40 v_1: \\
\frac{1}{v_1} = \frac{1}{-15} + \frac{1}{40} \\
v 1 : v 1 1 = − 15 1 + 40 1 Calculate the right-hand side:
1 v 1 = − 8 + 3 120 = − 5 120 = − 1 24 v 1 = − 24 \\
\frac{1}{v_1} = \frac{-8 + 3}{120} = \frac{-5}{120} = \frac{-1}{24} \\
v_1 = -24 v 1 1 = 120 − 8 + 3 = 120 − 5 = 24 − 1 v 1 = − 24 cm
Now, the object is displaced 20 cm towards the mirror, so the new object distance is:
u 2 = − 40 + 20 = − 20 \\
u_2 = -40 + 20 = -20 u 2 = − 40 + 20 = − 20 cm
Using the mirror formula again:
1 − 15 = 1 v 2 + 1 − 20 \\
\frac{1}{-15} = \frac{1}{v_2} + \frac{1}{-20} \\
− 15 1 = v 2 1 + − 20 1 Rearrange to solve for
v 2 : 1 v 2 = 1 − 15 + 1 20 v_2: \\
\frac{1}{v_2} = \frac{1}{-15} + \frac{1}{20} \\
v 2 : v 2 1 = − 15 1 + 20 1 Calculate the right-hand side:
1 v 2 = − 4 + 3 60 = − 1 60 v 2 = − 60 \\
\frac{1}{v_2} = \frac{-4 + 3}{60} = \frac{-1}{60} \\
v_2 = -60 v 2 1 = 60 − 4 + 3 = 60 − 1 v 2 = − 60 cm
The displacement of the image is the difference between the final and initial image distances:
Δ v = v 2 − v 1 = − 60 − ( − 24 ) = − 60 + 24 = − 36 \\
\Delta v = v_2 - v_1 = -60 - (-24) = -60 + 24 = -36 Δ v = v 2 − v 1 = − 60 − ( − 24 ) = − 60 + 24 = − 36 cm
The negative sign indicates that the image moves away from the mirror.
Therefore, the displacement of the image is
cm away from the mirror, which corresponds to Option 2.