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Question:
An erect image four times the size of the object is obtained with a concave mirror of radius of curvature 40 cm what is the position of the object?
Answer:

Using the mirror equation:

1/f = 1/v + 1/u

Where:

  • f is the focal length of the mirror.
  • v is the image distance.
  • u is the object distance.

The image is four times the size of the object, which means the magnification (M) is 4. The magnification can be calculated as:

M = -v/u

Since the image is erect, the magnification is positive. So, M = 4.

To find the focal length of the concave mirror using the given radius of curvature (R).

The focal length (f) and the radius of curvature (R) are related by the following equation for a concave mirror:

f = R/2

Given R = 40 cm, you can find:

f = 40 cm / 2 = 20 cm

Now that you have the focal length, you can use the magnification equation to find the object distance (u):

M = -v/u

4 = -v/u

v = -4u

Now, let's substitute the values we know:

f = 20 cm M = 4

1/f = 1/v + 1/u

1/20 = 1/(-4u) + 1/u

Now, solve for u:

1/20 = (-1/4u) + (1/u)

To simplify this equation, find a common denominator:

1/20 = (-u + 4u) / (4u2 )

1/20 = (3u) / (4u2 )

Now, cross-multiply:

4u^2 = 20 * 3u

4u2 = 60u

Divide both sides by 4:

u2 = 15u

Now, divide both sides by u:

u = 15

So, the object distance (u) is 15 cm. The object is located 15 cm in front of the concave mirror.

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