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CBSE Class 10 Science Question 15 of 31

Light - Reflection and Refraction — Question 15

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Question 15

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Answer

Given,

Object distance (u) = -27 cm [∵ object is placed infront of mirror so object distance will be negative]

Object height (h) = 7 cm

Image distance (v) = ?

Image height (h) = ?

f = -18 cm [∵ focal length of concave mirror is negative by convention]

According to the mirror formula,

1v\dfrac{1}{v} + 1u\dfrac{1}{u} = 1f\dfrac{1}{f}

Substituting the values we get,

1v+1−27=1−18⇒1v=127−118⇒1v=2−354⇒1v=−154⇒v=−54 cm\dfrac{1}{v} + \dfrac{1}{-27} = \dfrac{1}{-18} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 27} - \dfrac{1}{18} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{2-3}{54} \\[1em] \Rightarrow \dfrac{1}{v} = - \dfrac{1}{54} \\[1em] \Rightarrow v = - 54\text{ cm}

∴ The screen should be placed at a distance of 54 cm in front of the mirror on the object side.

Magnification=−vu=height of imageheight of object−−54−27=height of image7⇒height of image=−54×727⇒height of image=−14 cm\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \\[1em] -\dfrac{-54}{-27} = \dfrac{\text{height of image}}{7} \\[1em] \Rightarrow \text{height of image} = -\dfrac{54 \times 7}{27} \\[1em] \Rightarrow \text{height of image} = -14 \text{ cm}

∴ Height of image is 14 cm and image is real, inverted and magnified.