Your answer is partially correct. Try again. In the figure, an object is placed

Question

Your answer is partially correct. Try again. In the figure, an object is placed in front of a converging lens at a distance equal to twice the focal length f_1 of the lens. On the other side of the lens is a concave mirror of focal length f_2 separated from the lens by a distance 2(f_1 + f_2)- Light from the object passes rightward through the lens, reflects from the mirror, passes leftward through the lens, and forms a final image of the object. Take f_1 - 4.8 cm and f_2 - 2.8 cm. What are (a) the distance between the lens and the final image and (b) the overall lateral magnification M of the object? Is the image (c) real or virtual (if it is virtual, it requires someone looking through the lens toward the mirror), (d) to the left or right of the lens, and (e) inverted or noninverted relative to the object?

Explanation / Answer

We solve this problem in three steps:

For the lens: p1 = object distance, q1 = image distance and f1 is the focal length

f1 = 4.8 cm, p1 = 2*f1 = 9.6 cm

Use 1/p + 1/q = 1/f

1/9.6 + 1/q1 = 1/4.8

Solving, we get q1 = 9.6 cm

The magnification is M1 = - q1/p1 = - 9.6/9.6 = -1

Because M1 is negative, the image is inverted.

q1 is positive means the image is behind the lens. Its location with respect to the mirror is:

2(f1 + f2) - 9.6 = p2, the object of the mirror

So p2 = 2(4.8 + 2.8) - 9.6 = 5.6 cm

f2 = 2.8 cm so

1/5.6 + 1/q2 = 1/2.8

Solving, we get q2 = 5.6 cm

This image is inverted with respect of the inverted image of the lens which is the object for this second part of the solution.

q2 forms the object of the lens. The 'front' of the lens is on the right. The object's location is

p3 = 2(f1 + f2) - 5.6 = 2(4.8 + 2.8) - 5.6

p3 = 9.6

f1 = 4.8 so

1/9.6 + 1/q3 = 1/4.8 which gives

q3 = 9.6 cm

Because q3 is positive, it is at the left of the lens. And this is the distance required for
part (a).

The magnification is M3 = - q3/p3 = - 9.6/9.6 = -1

Because M is negative, the image is inverted with respect to its object at q3.

The overall magnification is M = M1*M2*M3 = (-1) * (-1) * (-1) = -1

So the final image is inverted with respect to the original object and the

magnification is 1.

I think others are already solve.

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