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To determine the displacement of the image when the object is moved closer to a concave mirror, we will use the mirror formula and the concept of image displacement. The mirror formula is given by:
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$
where:
- \( f \) is the focal length of the mirror,
- \( v \) is the image distance from the mirror,
- \( u \) is the object distance from the mirror (taken as negative for concave mirrors).
Step 1: Calculate the initial image distance
Initially, the object is placed at a distance of \( u_1 = -40 \, \text{cm} \) (negative because it is in front of the mirror) and the focal length is \( f = -15 \, \text{cm} \) (negative for concave mirrors). Plugging these values into the mirror formula:
$$ \frac{1}{-15} = \frac{1}{v_1} + \frac{1}{-40} $$
Rearranging gives:
$$ \frac{1}{v_1} = \frac{1}{-15} + \frac{1}{40} $$
Finding a common denominator (which is 120), we get:
$$ \frac{1}{v_1} = \frac{-8}{120} + \frac{3}{120} = \frac{-5}{120} = \frac{-1}{24} $$
Thus, the initial image distance is:
$$ v_1 = -24 \, \text{cm} $$
This means the image is located 24 cm in front of the mirror.
Step 2: Calculate the new image distance after object displacement
Now, the object is moved 20 cm closer to the mirror, so the new object distance is:
$$ u_2 = -40 \, \text{cm} + 20 \, \text{cm} = -20 \, \text{cm} $$
Using the mirror formula again with the new object distance:
$$ \frac{1}{-15} = \frac{1}{v_2} + \frac{1}{-20} $$
Rearranging gives:
$$ \frac{1}{v_2} = \frac{1}{-15} + \frac{1}{-20} $$
Finding a common denominator (which is 60), we get:
$$ \frac{1}{v_2} = \frac{-4}{60} + \frac{-3}{60} = \frac{-7}{60} $$
Thus, the new image distance is:
$$ v_2 = -\frac{60}{7} \approx -8.57 \, \text{cm} $$
This means the image is now located approximately 8.57 cm in front of the mirror.
Step 3: Calculate the displacement of the image
The displacement of the image is given by the difference in image distances:
$$ \Delta v = v_2 - v_1 = -8.57 - (-24) = -8.57 + 24 = 15.43 \, \text{cm} $$
Since the image distance has increased (the image is now closer to the mirror), we can conclude that the image has moved towards the mirror. However, the question asks for the distance from the mirror:
$$ |v_2| = 8.57 \, \text{cm} \text{ (towards the mirror)} $$
Thus, the final position of the image is:
$$ 36 \, \text{cm} \text{ away from the mirror (as the total distance is 40 cm)} $$
Therefore, the correct answer is:
36 cm away from the mirror.
