(a) State the mirror formula and define each term in it.
(b) A concave mirror has a focal length of 25 cm. An object is placed 40 cm in front of it. Find the position of the image. Is the image real or virtual?
(c) Now find the magnification. If the object is 2 cm tall, what is the height of the image? State whether the image is erect or inverted.
Generated by claude-sonnet-4-6 · 2026-06-26 01:13 · grounding rag
Model Answer
(a) Mirror Formula:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
- u = object distance (from pole of mirror)
- v = image distance (from pole of mirror)
- f = focal length (distance of principal focus from pole)
(b) Finding image position:
Given: f = –25 cm (concave mirror), u = –40 cm
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-25} - \frac{1}{-40} = -\frac{1}{25} + \frac{1}{40} = \frac{-8+5}{200} = \frac{-3}{200}$$
$$v = -\frac{200}{3} \approx -66.7 \text{ cm}$$
The image is formed 66.7 cm in front of the mirror. Since v is negative, the image is real.
(c) Magnification and image height:
$$m = -\frac{v}{u} = -\frac{(-66.7)}{(-40)} = -1.67$$
Height of image: $h' = m \times h = -1.67 \times 2 = -3.34 \text{ cm}$
The image is 3.34 cm tall. The negative sign indicates the image is inverted.
Source: Chapter 9, Section 9.2.4
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Explanation
- Always apply sign convention: distances in front of the mirror are negative; concave mirror focal length is negative.
- A negative value of v confirms a real image (formed in front of the mirror).
- A negative magnification confirms the image is inverted; magnitude > 1 means it is enlarged.
- Examiners award marks separately for formula, substitution, answer, and nature of image — show each step clearly.