Question

Two objects \(A\) and \(B\)are placed at \(15\, cm\) and \(25\, cm\) from the pole in front of a concave mirros having radius of curvature \(40\, cm\). The distance between images formed by the mirror is:

• A. \(40 \,cm\)
• B. \(160 \,cm\)
• C. \(60 \,cm\)
• D. \(100\, cm\)

Hint:

Remember the sign conventions for concave mirrors. The mirror formula is essential for solving problems involving image formation.

Answer 1

The Correct Option is B

Step 1: Determine the Focal Length

The focal length (\(f\)) of a concave mirror is half of its radius of curvature (\(R\)):

\[ f = \frac{R}{2} = \frac{40}{2} = -20 \, \text{cm} \]

The focal length is negative for a concave mirror.

Step 2: Use the Mirror Formula for Object A

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

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

For object A, \(u_1 = -15 \, \text{cm}\) (negative because the object is in front of the mirror).

\[ \frac{1}{v_1} + \frac{1}{-15} = \frac{1}{-20} \] \[ \frac{1}{v_1} = \frac{1}{15} - \frac{1}{20} = \frac{20 - 15}{300} = \frac{5}{300} = \frac{1}{60} \] \[ v_1 = 60 \, \text{cm} \]

Step 3: Use the Mirror Formula for Object B

For object B, \(u_2 = -25 \, \text{cm}\):

\[ \frac{1}{v_2} + \frac{1}{-25} = \frac{1}{-20} \] \[ \frac{1}{v_2} = \frac{1}{25} - \frac{1}{20} = \frac{20 - 25}{500} = \frac{-5}{500} = \frac{-1}{100} \] \[ v_2 = -100 \, \text{cm} \]

Step 4: Calculate the Distance Between the Images

The image of A is formed at \(v_1 = 60 \, \text{cm}\) (positive, so it’s a real image formed in front of the mirror). The image of B is formed at \(v_2 = -100 \, \text{cm}\) (negative, so it’s a virtual image formed behind the mirror). The distance (\(d\)) between the images is:

\[ d = |v_1| + |v_2| = 60 + 100 = 160 \, \text{cm} \]

Conclusion:

The distance between the images is 160 cm (Option 2).