Question

An object is placed at a distance of 10 cm from a convex mirror of focal lenen
15 cm. Find the position and nature of the image.


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Answer 1

Answer :-

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Mirror Formula and Sign Convention for Convex Mirror has been used. We know that for Convex Mirror, the focal length and distance of image is taken positive is taken positive since they are formed at right side that is behind mirror. Whereas the distance of object is taken as negative since its in front of mirror and and at left side of mirror. Using this concept,

let's do it !!

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★ Formula Used :-

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

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★ Solution :-

Given,

» Distance of object from mirror = u = - 10 cm

» Focal Length of the mirror = f = 15 cm

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~ For the Distance of image from mirror :-

• Let the distance of image from the mirror be 'v'.

Then,

$\\\;\;\;\;\;\;\sf{:\Longrightarrow\;\;\;\dfrac{1}{f}\;=\;\dfrac{1}{v}\;+\;\dfrac{1}{u}}$

By applying values, we get,

$\\\;\;\;\;\;\;\sf{:\Longrightarrow\;\;\;\dfrac{1}{15}\;=\;\dfrac{1}{v}\;+\;\dfrac{1}{(-\:10)}}$

$\\\;\;\;\;\;\;\sf{:\Longrightarrow\;\;\;\dfrac{1}{15}\;=\;\dfrac{1}{v}\;+\;\dfrac{(-\:1)}{10}}$

$\\\;\;\;\;\;\;\sf{:\Longrightarrow\;\;\;\dfrac{1}{v}\;=\;\dfrac{1}{15}\;-\;\dfrac{(-\:1)}{10}}$

$\\\;\;\;\;\;\;\sf{:\Longrightarrow\;\;\;\dfrac{1}{v}\;=\;\dfrac{1}{15}\;+\;\dfrac{1}{10}}$

Taking the LCM, we get,

$\\\;\;\;\;\;\;\sf{:\Longrightarrow\;\;\;\dfrac{1}{v}\;=\;\dfrac{2\;+\;3}{30}}$

$\\\;\;\;\;\;\;\sf{:\Longrightarrow\;\;\;\dfrac{1}{v}\;=\;\dfrac{5}{30}}$

$\\\;\;\;\;\;\;\sf{:\Longrightarrow\;\;\;\dfrac{1}{v}\;=\;\dfrac{\cancel{5}}{\cancel{30}}}$

$\\\;\;\;\;\;\;\sf{:\Longrightarrow\;\;\;\dfrac{1}{v}\;=\;\dfrac{1}{6}}$

By taking the reciprocal, we get,

$\\\;\;\;\;\;\;\bf{:\Longrightarrow\;\;\;v\;=\;6\;\;cm}$

$\\\;\large{\underline{\underline{\rm{Thus,\;postion\;of\;image\;from\;mirror\;is\;\;\boxed{\bf{6\;\;cm}}}}}}$

Now we know that its a convex mirror, then image will be formed behind the mirror. This means image is Virtual, erect and diminished.

$\\\;\underline{\boxed{\tt{Nature\;\;of\;\;Image\;\;=\;\bf{Virtual,\;erect\;and\;diminished}}}}$

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★ More to know :-

Now let us find the magnification of the image.

Its denoted (m). Then,

Magnification is the ratio of Height of Image (hi) by Height of Object (h). Its also given as the ratio of Position of Image (v) by Position of Object (u).

$\\\;\sf{\leadsto\;\;Magnification_{(Mirror)},\;m\;=\;\dfrac{h_{i}}{h}\;=\;\dfrac{-\:(v)}{u}}$

$\\\;\sf{\leadsto\;\;Magnification_{(Mirror)},\;m\;=\;\dfrac{-\:(6)}{-\:(10)}}$

$\\\;\sf{\leadsto\;\;Magnification_{(Mirror)},\;m\;=\;\dfrac{(6)}{(10)}}$

$\\\;\bf{\leadsto\;\;Magnification_{(Mirror)},\;m\;=\;\underline{\underline{0.6}}}$

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★ Additional Formulas :-

Lens Formula is given as ::

$\\\;\tt{\mapsto\;\;\;\dfrac{1}{f}\;\;=\;\;\dfrac{1}{v}\;\;-\;\;\dfrac{1}{u}}$

$\\\;\tt{\mapsto\;\;\;Power,\;P\;\;=\;\;\dfrac{1}{f}}$

$\\\;\tt{\mapsto\;\;\;Power\;\;of\;\;Combined\;\;Lenses,\;P\;\;=\;\;p_{1}\;+\;p_{2}\;+\;p_{3}\;+\;........\;+\;p_{n}}$

$\\\;\tt{\mapsto\;\;\;Magnification_{(Lens)},\;m\;=\;\dfrac{h_{i}}{h}\;=\;\dfrac{v}{u}}$

Answer 2

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