Question

Study the data given below showing the focal length of three concave mirrors A, B and C and the respective distances of objects placed in front of the mirrors:

Case	Mirror	Focal Length (cm)	Object Distance (cm)
1	A	20	45
2	B	15	30
3	C	30	20

(i) In which one of the above cases the mirror will form a diminished image of the object? Justify your answer.
(ii) List two properties of the image formed in case 2.
(iii) (A) What is the nature and size of the image formed by mirror C? Draw ray diagram to justify your answer.

OR

(iii) (B) An object is placed at a distance of 18 cm from the pole of a concave mirror of focal length 12 cm. Find the position of the image formed in this case.

Answer 1

(i). Case 1 (Mirror A): The object distance (45 cm) is beyond the center of curvature (C = 2f = 40 cm).
When the object is beyond C, the image formed is diminished, inverted, and between f and C.

(ii). List two properties of the image formed in case 2.
Solution: In Case 2 (Mirror B): The object distance (30 cm) is at the center of curvature (C = 2f = 30 cm).
Properties of the image:
• The image is real and inverted.
• The image is of the same size as the object.

(iii)A In Case 3 (Mirror C): The object distance (20 cm) is less than the focal length(30 cm).
Nature of Image: Virtual, erect, and magnified.
Ray Diagram: Draw a ray originating from the object parallel to the principal axis and refracting through the focal point. Another ray passes through the center of curvature, and the virtual rays appear to meet behind the mirror.

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(ii) Using the mirror formula: 
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] 
\[ \frac{1}{12} = \frac{1}{v} - \frac{1}{-18} \] 
\[ \frac{1}{v} = \frac{1}{12} + \frac{1}{18} = \frac{3 + 2}{36} = \frac{5}{36} \] 
\[ v = 7.2 \, \text{cm} \, (\text{Image is real and inverted at } 7.2 \, \text{cm from the mirror}). \]