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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:

CaseMirrorFocal Length (cm)Object Distance (cm)
1A2045
2B1530
3C3020

(a) In which one of the above cases the mirror will form a diminished image of the object? Justify your answer.

(b) List two properties of the image formed in case 2.

(c) What is the nature and size of the image formed by mirror C? Draw ray diagram to justify your answer.

OR

(c) 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.

Reflection of Light

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Answer

(a) Case 1 (Mirror A) will form a diminished image.

Explanation : The object (45 cm) is beyond C (40 cm) and for a concave mirror an object beyond C produces a real, inverted and diminished image located between f and C.

(b) Two properties of the image formed in case 2 (Mirror B) where object is at C :

  1. The image is real and inverted.
  2. The image is of the same size as the object and is formed at C (30 cm) on the other side of the mirror.

(c) For mirror C (f = 30 cm, object at 20 cm) :

  • Nature of image : Virtual and erect (since the object lies between pole and focus).
  • Size of image: Magnified (appears larger than the object).
  • Where formed : Behind the mirror (cannot be projected on a screen).

The ray diagram of the above arrangement is shown below :

What is the nature and size of the image formed by mirror C? Draw ray diagram to justify your answer.CBSE 2026 Science Class 10 Sample Question Paper Solved.

OR

(c)

Given,

  • Object distance (u\text u) = -18 cm
  • Focal length of the mirror (f\text f) = -12 cm

Here, negative sign indicates that object is in front of the mirror and parallel rays after reflection converges in front of the mirror.

Let, image distance be 'v\text v'.

From mirror formula,

1f=1u+1v1v=1f1u=1(12)1(18)=112+118=23361v=136v=36 cm\dfrac{1}{\text f} = \dfrac{1}{\text u} + \dfrac{1}{\text v} \\[1em] \Rightarrow \dfrac{1}{\text v} = \dfrac{1}{\text f} - \dfrac{1}{\text u} \\[1em] = \dfrac{1}{(-12)} - \dfrac{1}{(-18)} \\[1em] = -\dfrac{1}{12} + \dfrac{1}{18} \\[1em] = \dfrac{2 - 3}{36} \\[1em] \Rightarrow \dfrac{1}{\text v} = \dfrac{-1}{36} \\[1em] \Rightarrow \text v = -36 \text { cm}

Hence, the image is formed at a distance of 36 cm in front of the mirror.

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