Question

Size of image of an object by a mirror having focal length of 40cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror.

Answer 1

Question :-

Size of image of an object by a mirror having focal length of 40cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror.

Solution :-

Using magnification formula

$\sf{m = - \frac{v}{u} = \frac{1}{3} } \\ \\ \implies \sf{v = - \frac{u}{ 3} }$

From mirror formula

$\sf:\longmapsto{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }$

From above two equations

$\sf:\longmapsto{ \frac{3}{ - u} + \frac{1}{ - u} = \frac{1}{40} } \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{ \frac{ - 3u + ( - 1u)}{u {}^{2} } = \frac{1}{40} } \\ \\ \sf:\longmapsto{ \frac{ - 4u}{u {}^{2} } = \frac{1}{40} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{ - 4u \times 40 = u {}^{2} } \: \: \: \: \: \\ \\ \sf:\longmapsto{ u {}^{2} = - 160u } \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{ \frac{u {}^{2} }{u} = - 160} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{u = - 160} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• u = -160cm

So, the object is placed at a distance of 80cm in front of the mirror.

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size of image of an object by a mirror having a focal length of 20cm is observed to be reduced to 1/3rd to its size. At what distance the object has been placed from the mirror? what is the nature of the image and the mirror?.

https://brainly.in/question/2642068

Answer 2

SOLUTION :-

Using Magnification Formula,

$\sf \star \fbox \orange{m = - \frac{v}{u}} = \frac{1}{3}$

$\sf →v = -\frac{u}{3}$

Now, from Mirror Formula,

$\sf \star \fbox \orange {\frac{1}{v} + \frac{1}{u} = \frac{1}{f}}$

Putting value of $\sf v$ in the mirror formula,

$\sf → \frac{3}{ - u} + \frac{1}{ - u} = \frac{1}{40}$

$\sf → \frac{- 3u + ( - u)}{ {u}^{2} } = \frac{1}{40}$

$\sf → \frac{-4u}{ {u}^{2} } = \frac{1}{40}$

Cross multiplying,

$\sf → - 4u \times 40 = {u}^{2}$

$\sf → -160u = {u}^{2}$

$\sf → \frac{ {u}^{2} }{u} = 160$

$\sf \therefore \fbox{u = - 160 \: cm}$

Hence, the object is placed at the distance of 160 cm in the front of mirror.