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An object is placed 40cm from a converging lens of focal length 25cm . if the real image which is 5cm is formed, calculate the height of the object?



User Redzo
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Using the lens equation, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.

We are given that f = 25 cm, u = -40 cm (since the object is placed in front of the lens), and v = -5 cm (since the image is real).

Substituting these values into the lens equation, we get:

1/25 = 1/-5 - 1/-40

Simplifying this equation yields:

1/25 = 7/200

Multiplying both sides by 200, we get:

8 = v

Therefore, the image distance is v = -8 cm.

Using the magnification equation, m = -v/u, where m is the magnification, we can find the magnification of the image:

m = -v/u = -(-8)/(-40) = 1/5

Since the magnification is positive, the image is upright.

The height of the object, h1, is related to the height of the image, h2, by the magnification equation:

m = h2/h1

Substituting the known values of m and h2, we get:

1/5 = 5/h1

Solving for h1, we get:

h1 = 25 cm

Therefore, the height of the object is 25 cm.
User Abdeen
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