Question

The focal length of a concave mirror is f and the distance from the object to the principle focus is x. The ratio of the size of the image to the size of the object is

a) $\sf\dfrac{f+x}{f}$

b) $\sf\dfrac{f}{x}$

c) [tex}\sf\sqrt{\dfrac{f}{x}}[/tex]​

Answer 1

★ Concept :-

Here the concept of sign convention of Concave Mirror has been used. We see that we are given the focal length of the concave mirror. Even we are given the distance of object from principal focus. So firstly we can find the total distance of the object from mirror. This will give us a relationship between the focal length and the distance of image from the mirror. Then we can use the formula of magnification to find our answer.

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★ Formula Used :-

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

$\;\boxed{\sf{\pink{m\;=\;\dfrac{h_{0}}{h}\;=\;\dfrac{-v}{u}}}}$

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★ Solution :-

Given,

» Type of spherical mirror = Concave Mirror

» Focal Length of the mirror = f

» Distance of object from principal focus = x

• Let the distance of object from mirror be u

• Let the distance of image from mirror be v

According to the new sign convention, we know that distance of object from mirror is taken as negative. So, u = -u

And even the we are given the focal length of mirror and according to new sign convention for concave mirror, focal length is taken as negative. So, f = -f

Now we are given that object is placed beyond the principal focus. Since, the distance between principal focus and object is already given to us.

So total distance of object from mirror will be the sum of this distance and focal length. So,

→ u = -(f + x)

Now let's apply this in the Mirror Formula.

$\;\;\tt{\rightarrow\;\;\dfrac{1}{f}\;=\;\dfrac{1}{v}\;+\;\dfrac{1}{u}}$

$\;\;\tt{\rightarrow\;\;\dfrac{1}{f}\;=\;\dfrac{1}{v}\;+\;\dfrac{1}{-(f\:+\;x)}}$

$\;\;\tt{\rightarrow\;\;\dfrac{1}{-(f)}\;=\;\dfrac{1}{v}\;+\;\dfrac{1}{-(f\:+\;x)}}$

$\;\;\tt{\rightarrow\;\;\dfrac{-1}{f}\;=\;\dfrac{1}{v}\;-\;\dfrac{1}{(f\:+\;x)}}$

On transposing 1/u to other side, we get

$\;\;\tt{\rightarrow\;\;\dfrac{1}{v}\;=\;\dfrac{-1}{f}\;+\;\dfrac{1}{(f\:+\;x)}}$

On taking the L.C.M. we get,

$\;\;\tt{\rightarrow\;\;\dfrac{1}{v}\;=\;\dfrac{-(f\:+\:x)\;+\;f}{f(f\:+\;x)}}$

$\;\;\tt{\rightarrow\;\;\dfrac{1}{v}\;=\;\dfrac{-f\:-\:x\;+\;f}{f\:\times\:(f\:+\;x)}}$

$\;\;\tt{\rightarrow\;\;\dfrac{1}{v}\;=\;\dfrac{-\:x}{f\:\times\:(f\:+\;x)}}$

Now taking -ve sign as common, we get

$\;\;\tt{\rightarrow\;\;\dfrac{1}{v}\;=\;-\dfrac{x}{f\:\times\:(f\:+\;x)}}$

Now taking the reciprocal at both sides, we get

$\;\;\bf{\rightarrow\;\;\red{v\;=\;-\dfrac{f\:\times\:(f\:+\;x)}{x}}}$

Since the distance of image from mirror is -ve, this means image is real and inverted.

We have the formula that,

$\;\tt{\rightarrow\;\;m\;=\;\dfrac{h_{0}}{h}\;=\;\dfrac{-v}{u}}$

• Here m = magnification of mirror

• Here h｡= Height of image

• Here h = Height of object

Clearly from the formula, we see that

$\;\tt{\rightarrow\;\dfrac{h_{0}}{h}\;=\;\dfrac{-v}{u}}$

This means ratio of height of image by ratio of height of object is equal to the ratio of distance of image from mirror by distance of object from mirror.

On apply values, we get

$\;\tt{\rightarrow\;\dfrac{h_{0}}{h}\;=\;\dfrac{-\bigg(-\dfrac{f\:\times\:(f\:+\;x)}{x}\bigg)}{-(f\:+\:x)}}$

Cancelling the -ve sign from numerator, we get

$\;\tt{\rightarrow\;\dfrac{h_{0}}{h}\;=\;\dfrac{\bigg(\dfrac{f\:\times\:(f\:+\;x)}{x}\bigg)}{-(f\:+\:x)}}$

$\;\tt{\rightarrow\;\dfrac{h_{0}}{h}\;=\;\dfrac{f\:\times\:(f\:+\;x)}{-x\:\times\:(f\:+\:x)}}$

Cancellating (f + x) from numerator and denominator, we get

$\;\tt{\rightarrow\;\dfrac{h_{0}}{h}\;=\;\dfrac{f}{-x}}$

Since ratio of heights is always positive, thus this can be written as,

$\;\bf{\rightarrow\;\green{\dfrac{h_{0}}{h}\;=\;\dfrac{f}{x}}}$

(-ve sign in ratio of heights means that the image is inverted)

This is the required answer.

So the correct option is b) f/x

$\;\underline{\boxed{\tt{Required\;\:ratio\;\:of\;\:Heights\;=\;\bf{\purple{\dfrac{f}{x}}}}}}$

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★ More to know :-

• About Lenses ::

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

• This is the Lens Formula

$\;\sf{\Longrightarrow\;\;m\;=\;\dfrac{h_{0}}{h}\;=\;\dfrac{v}{u}}$

• This is the formula for magnification of image by lens.

$\;\sf{\Longrightarrow\;\;Power\;=\;\dfrac{1}{f}}$

• This is formula for the power of yhe lens.

Answer 2

★ Question Given :

• The focal length of a concave mirror is f and the distance from the object to the principle focus is x. The ratio of the size of the image to the size of the object is

• a) f + x / f

• b) f / x

• c) fx

★ Required Solution :

✯ Values Given to us :

• ➷ Focal length of a concave mirror is f

• ➷ The distance from the object to the principle focus is x.

✯ Formulas used here :

• ➷ Mirror Formula = 1/v + 1/u = 1/f

• ➷ Magnification formula = - v/u

★ Concept behind :

• ➷ Basically this Question is based on sign convention

✯ Putting Values in formula :

• ➷ 1/v - 1/f+x = -1/f

• ➷ 1/v = -1/f + 1/f+x

• ➷ 1/v = p / (f + x) × f

• ➷ v = (f + x) × f / x

✯ Using Magnification Formula :

• ➴ - v/u

• ➴ - (f +x) × f / x × (f + x)

✯ After solving further :

• ➴ - f / x

• Can Also be Written as :

• ➴ f / x

• ✰ f / x ✰

• ➴ The ratio of the size of the image to the size of the object is f / x

➴ Option (b) is correct f / x ✰

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★ Knowledge Booster ★

• ➼ - ve sign indicates inverted image !

• ➼ -ve sign indicates image is real !

• ➼ f indicates focal length

• ➼ u is the Object distance.

• ➼ v is the Image distance.

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