Question

c) An object of length 4 cm is placed in front of a
concave mirror at a distance 30 cm. The focal length
of the mirror is 15 cm.
i) Where will the image form?
ii) What will be the length of the image?​

Answer 1

Explanation:

The relationship between the radius of curvature and focal length is that the radius of curvature is twice the focal length. In this case, focal length of the concave mirror is given as 15 cm. Hence, the radius of curvature is 15×2=30cm

Answer 2

Given:

• Height of image, $\sf h_{i}$ = 4 cm
• Object distance, u = - 30 cm
• Focal length, f = - 15 cm

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To find:

1. Where will the image form?
2. What will be the length of the image?

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Solution:

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$\bigstar\:{\underline{\sf Using\:Mirror\:Formula\::}}$

$\star\;{\boxed{\sf{\pink{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}}$

$:\implies\sf \dfrac{1}{-15} = \dfrac{1}{v} + \dfrac{1}{-30}$

$:\implies\sf \dfrac{1}{v} = \dfrac{1}{-15} + \dfrac{1}{30}$

$:\implies\sf \dfrac{1}{v} = \dfrac{- 2 + 1}{30}$

$:\implies\sf \dfrac{1}{v} = \dfrac{- 1}{30}$

$:\implies{\underline{\boxed{\frak{\purple{v = -30\:cm}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Distance\;of\;image\;from\:is\; \bf{30\:cm}.}}}$

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$\bigstar\:{\underline{\sf Now,\: from\:Magnification\:Formula\::}}$

$\star\;{\boxed{\sf{\pink{m = \dfrac{h_i}{h_o} = \dfrac{-v}{u}}}}}$

$:\implies\sf \dfrac{h_i}{4} = \dfrac{- \cancel{(-30)}}{ \cancel{-30}}$

$:\implies\sf \dfrac{h_i}{4} = - 1$

$:\implies\sf h_i = -1 \times 4$

$:\implies{\underline{\boxed{\frak{\purple{h_i = -4\:cm}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Length\;of\;image\;is\; \bf{4\:cm}.}}}$

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$\bigstar\:{\underline{\sf Nature\:of\:image\::}}$

• At centre of curvature (C).
• Real and inverted.