A mirror makes an image 3 times larger image of an object whose focal length is 30 cm. Calculate the possible positions of the object.



The magnification produced by a mirror is given by the formula:

\[
m = \frac{-v}{u}
\]

where:
- \(m\) is the magnification,
- \(v\) is the image distance from the mirror,
- \(u\) is the object distance from the mirror.

For this problem:
- The magnification \(m = 3\) (since the image is 3 times larger),
- The focal length of the mirror \(f = 30 \, \text{cm}\).

Also, for a mirror, the mirror formula relates the object distance \(u\), the image distance \(v\), and the focal length \(f\):

\[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
\]

### Step 1: Using magnification formula

Since magnification \(m = 3\):

\[
3 = \frac{-v}{u} \implies v = -3u
\]

### Step 2: Substitute into the mirror formula

Now, substitute \(v = -3u\) into the mirror formula:

\[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
\]

\[
\frac{1}{30} = \frac{1}{-3u} + \frac{1}{u}
\]

### Step 3: Solve for \(u\)

\[
\frac{1}{30} = \frac{-1}{3u} + \frac{1}{u}
\]

Find a common denominator on the right-hand side:

\[
\frac{1}{30} = \frac{-1 + 3}{3u} = \frac{2}{3u}
\]

Now, solve for \(u\):

\[
\frac{1}{30} = \frac{2}{3u}
\]

Multiply both sides by \(3u\):

\[
3u \cdot \frac{1}{30} = 2 \implies \frac{u}{10} = 2 \implies u = 20 \, \text{cm}
\]

### Step 4: Calculate \(v\)

Since \(v = -3u\), substitute \(u = 20 \, \text{cm}\):

\[
v = -3 \times 20 = -60 \, \text{cm}
\]

### Possible object positions

The object should be placed **20 cm** in front of the mirror, and the image will be **60 cm** behind the mirror.

#### Do you need more details or have any questions?

Here are five related questions to extend your understanding:

1. What kind of mirror could produce a magnified virtual image?
2. How does the nature of the image (real or virtual) change with different object positions in concave mirrors?
3. What happens if the object is placed exactly at the focal length of the mirror?
4. Can a convex mirror ever produce a magnified image?
5. How does magnification relate to the curvature of the mirror?

**Tip:** In concave mirrors, real images are inverted and magnified when the object is between the focal point and the center of curvature.

A mirror makes an image 3 times larger image of an object whose focal length is 30 cm. Calculate the possible positions of the object.

Learn how to calculate the object position for a mirror that creates an image three times larger with a focal length of 30 cm. This step-by-step solution uses the magnification formula and mirror formula to find the correct positions for the object and image. Suitable for students in Grades 9-12.

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