Reflection At Spherical Mirrors
An object is placed 30 cm in front of a concave mirror whose focal length is 20 cm. What is the magnification of the image that the mirror forms in this arrangement?
Select the correct option:
Solution
−2
A spherical mirror obeys the mirror equation v1+u1=f1, while the linear magnification follows m=−uv, both written in the Cartesian sign convention where distances measured against the incident light are negative. For a concave mirror the focal length is taken as negative, so f=−20 cm and the object distance is u=−30 cm. Rearranging gives v1=f1−u1=−201+301=−601, hence v=−60 cm. The magnification is m=−uv=−−30−60=−2, a real, inverted image twice the object size. Physically, rays diverging from the object converge after reflection from the concave surface, and the two standard construction rays—one travelling parallel to the principal axis then passing through the focus, and one heading toward the centre of curvature and retracing its path—intersect to fix the real image. The negative sign of m encodes the inversion, while its magnitude exceeding unity signals genuine enlargement of the object. The value +2 is wrong because it ignores the inversion implied by the negative sign. The value −0.5 is wrong as it inverts the ratio v/u. The value −3 is wrong since it would require an image distance of 90 cm. This is the standard NCERT treatment of image formation by concave mirrors. A plausibility check confirms that an object lying between the focus and the centre of curvature must yield a magnified inverted image, consistent with ∣m∣>1.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More reflection at spherical mirrors Practice Questions
About This Question
- Subject
- physics
- Chapter
- optics
- Topic
- reflection at spherical mirrors
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
−2
A spherical mirror obeys the mirror equation v1+u1=f1, while the linear magnification follows m=−uv, both written in the Cartesian sign convention where distances measured against the incident light are negative. For a concave mirror the focal length is taken as negative, so f=−20 cm and the object distance is u=−30 cm. Rearranging gives v1=f1−u1=−201+301=−601, hence v=−60 cm. The magnification is m=−uv=−−30−60=−2, a real, inverted image twice the object size. Physically, rays diverging from the object converge after reflection from the concave surface, and the two standard construction rays—one travelling parallel to the principal axis then passing through the focus, and one heading toward the centre of curvature and retracing its path—intersect to fix the real image. The negative sign of m encodes the inversion, while its magnitude exceeding unity signals genuine enlargement of the object. The value +2 is wrong because it ignores the inversion implied by the negative sign. The value −0.5 is wrong as it inverts the ratio v/u. The value −3 is wrong since it would require an image distance of 90 cm. This is the standard NCERT treatment of image formation by concave mirrors. A plausibility check confirms that an object lying between the focus and the centre of curvature must yield a magnified inverted image, consistent with ∣m∣>1.
This easy difficulty physics question is from the chapter optics, covering the topic of reflection at spherical mirrors. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse optics questions on RankGuru.