Question

If a 20 cm tall rabbit is playing at a distance of 30 cm in front of a concave mirror attached to its cage. What will be its height in the mirror if the focal length is 16 cm?​

Answer 1

Given:

• Height of object = ho = 20cm
• Object distance = u = -30cm
• Focal length = f = -16 cm

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Need to find:

• Height of image = Hi= ?

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Solution:

We know, the mirror formula:

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

$\implies \dfrac{1}{v} + \dfrac{1}{( - 30)} = \dfrac{1}{( - 16)}$

$\implies \: \dfrac{1}{v} = \dfrac{1}{30} - \dfrac{1}{16}$

$\implies \: \dfrac{1}{v} = \dfrac{8 - 15}{240}$

$\implies \: \dfrac{1}{v} = \dfrac{ - 7}{240}$

$\purple{\boxed{\bold{ \implies \: v = \dfrac{ - 270}{7}}}}$

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We know,

Magnification of mirror =

$\dfrac{hi}{ho} = \dfrac{ - v}{u}$

$\implies \: \dfrac{hi}{20} = - ( - \dfrac{240}{7} \times \dfrac{1}{ - 30} )$

$\implies \: \dfrac{hi}{20} = \dfrac{ - 8}{7}$

$\implies \: hi = \dfrac{ - 8}{7} \times 20 \: cm$

$\implies \: hi = \dfrac{ - 160}{7}$

$\red{\large{\bold{\boxed{ \implies \: hi \approx \: - 22.8cm}}}}$

∴ The height of the rabbit in the mirror is 22.8cm.

Here -ve sign indicates that the image is real and inverted.

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$\setlength{\unitlength}{25} \begin{picture}(10,10)\thicklines\qbezier(7,6)(8,3)(7,1) \put(1,3.5){\line(1,0){6.5}} \put(1,3.5){\line(1,0){6.5}}\put(4.5,3.5){\vector(0,1){1}}\put(4.5,4.5){\vector(1,0){1}}\put(4.5,4.5){\line(1,0){2.9}}\put(4.5,4.5){\line(1, - 1){2.8}}\put(7.4,4.5){\line( - 2, - 1){6.5}}\put(4.5,4.5){\vector(1, - 1){0.7}}\put(4.5,4.5){\vector(1, - 1){2}}\put(7.4,4.5){\vector( - 2, - 1){1}}\put(7.4,4.5){\vector( - 2, - 1){4}}\put(7.3,1.7){\line( - 1,0){5.5}}\put(1.8,3.5){\vector(0, - 1){1.8}}\put(7.3,1.7){\vector( - 1,0){3}}\put(4.4,3.1){ \tt B }\put(3,3.1){ \tt C }\put(1.75,3.75){ \tt B' }\put(1.75,1.2){ \tt A' }\put(5.25,3.1){ \tt F }\put(4.4,4.75){ \tt A }\put(7.5,4.7){ \tt L }\put(7.7,3.25){ \tt P }\put(7.4,1.5){ \tt TheDiamond }\put(3.125,3.5){\circle*{0.15}}\put(5.4,3.5){\circle*{0.15}}\put(7.1,5.7){\line(1, - 1){0.5}}\put(7.3,4.8){\line(1, - 1){0.5}}\put(7.4,4.1){\line(1, - 1){0.5}}\put(7.1,1.3){\line(1, - 1){0.5}}\put(7.4,2.4){\line(1, - 1){0.5}}\put(7.5,3){\line(1, - 1){0.5}}\end{picture}$

Here

• AB= Object
• A' B' = Image
• F = Focal Length
• C= Centre of Curvature

Answer 2

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