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A Lens Produces A Real Image That Is Twice As Large As The Object And Is Located 15.0 Cm From The Lens. (a) Determine the object distance.(b) Find the focal length of the lens.(c) What type of lens was used in this case?

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Answer:

Given data:

Magnification, m = 2

Object distance, u = ?

Image distance, v = -15.0 cm (as the image is real and on the opposite side of the lens)

Using the lens formula:

1/f = 1/v - 1/u

Since the image is real, v is negative. Substituting the values:

1/f = 1/-15 - 1/u

Multiplying both sides by -15u, we get:

-15u/f = u - 15

Rearranging, we get:

15u/f = 15 + u

Taking the reciprocal of both sides, we get:

f/15 = (u+15)/15u

Simplifying, we get:

f = 15u/(u+15) ------ (1)

We also know that the magnification, m = -v/u, since the image is real.

Substituting the given values:

2 = -v/u

v = -2u

Substituting v in terms of u in equation (1), we get:

f = -30u/(u-15) ------ (2)

Equations (1) and (2) represent two equations in two variables (u and f). Solving them simultaneously gives:

u = 30 cm

f = 45 cm

The sign of f is positive, indicating a convex lens.

Therefore, the answers are:

(a) Object distance, u = 30 cm

(b) Focal length, f = 45 cm

(c) Type of lens, convex lens.

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