Question 8 of 24intermediate🔧 ApplyNumerical3 marks

An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Correct Answer

The image is formed at 8.4 cm from the lens on the same side as the object. It is virtual, erect and diminished with a size of 1.8 cm.

If the object is moved further away from the lens, the image will shift closer to the focus (at 21 cm from the lens) and become smaller in size.

Exercise: EXERCISES | Q: 9.9 | (Page 29)
For More Understanding

Explanation

This is a standard numerical problem on concave lens image formation. The lens formula 1/v - 1/u = 1/f is used with proper sign conventions: u is negative (object distance), f is negative (concave lens). The negative value of v confirms a virtual image formed on the same side as the object. The positive magnification (less than 1) indicates an erect and diminished image. When the object moves further away, the image approaches the focal point as a limiting position.

Solution Steps

  1. Step 1: Given: Object size h = 3.0 cm, Object distance u = -14 cm, Focal length f = -21 cm (negative for concave lens)

  2. Step 2: Using lens formula: 1/v - 1/u = 1/f

  3. Step 3: Substituting values: 1/v - 1/(-14) = 1/(-21)

  4. Step 4: Solving: 1/v + 1/14 = -1/21, therefore 1/v = -1/21 - 1/14 = -5/42

  5. Step 5: Image distance v = -42/5 = -8.4 cm (negative sign indicates virtual image on same side as object)

  6. Step 6: Magnification m = v/u = (-8.4)/(-14) = 0.6 (positive and less than 1, so erect and diminished)

  7. Step 7: Image size h' = m × h = 0.6×3.00.6 \times 3.0 = 1.8 cm

  8. Step 8: When object moves further away, as u → -∞, the image distance v → f = -21 cm (approaches focus)

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