Question

An object is placed at 60.0 cm in front of a concave mirror of focal length 30 cm. At what distance from the mirror should a screen be placed to obtain a sharp image?

• A. 60.0 cm in front of the mirror
• B. 30.0 cm back of the mirror
• C. 90.0 cm in front of the mirror
• D. 90.0 cm back of the mirror

Hint:

Notice that the object is placed at \(60\text{ cm}\), which is exactly twice the focal length (\(2f = 2 \times 30 = 60\text{ cm}\)).
When an object is placed at the center of curvature (\(C\)), its real image is also formed at the center of curvature (\(C\)).
This allows you to identify the answer instantly without doing calculations.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the position of a screen to capture a sharp image, which is equivalent to finding the image distance (\(v\)) for a given object distance (\(u\)) and focal length (\(f\)) of a concave mirror.

Step 2: Key Formula or Approach:
The mirror formula is given by:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]

Step 3: Detailed Explanation:

• According to Cartesian sign conventions:
- Focal length of the concave mirror (\(f\)) = \(-30\text{ cm}\)
- Object distance (\(u\)) = \(-60\text{ cm}\)

• Substituting these values into the mirror formula:
\[ \frac{1}{v} + \frac{1}{-60} = \frac{1}{-30} \] \[ \frac{1}{v} - \frac{1}{60} = -\frac{1}{30} \]

• Solving for \(\frac{1}{v}\):
\[ \frac{1}{v} = -\frac{1}{30} + \frac{1}{60} \]

• Finding a common denominator:
\[ \frac{1}{v} = \frac{-2 + 1}{60} = -\frac{1}{60} \] \[ v = -60\text{ cm} \]

• The negative sign of \(v\) indicates that the image is real and formed in front of the mirror.

• Since real images can be captured on a screen, the screen must be placed at a distance of \(60.0\text{ cm}\) in front of the mirror.


Step 4: Final Answer:
The screen should be placed at 60.0 cm in front of the mirror.