Tag: problems on mirror and magnification formula

Questions Related to problems on mirror and magnification formula

A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placing

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. a concave mirror of suitable focal length

  2. a convex mirror of suitable focal length

  3. a convex lens of focal length less than 0.25 m

  4. a concave lens of suitable focal length

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Correct Option: C
Explanation:

Image can be formed on the screed if it is real. Real image of reduced size can be formed can be formed by a concave mirror or a convex lens.


The object is beyond $2f$. 

So let $u=2f+x$

And using lens formula we have

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

or

$\dfrac{1}{2f+x}+\dfrac{1}{v}=\dfrac{1}{f}$

or

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{2f+x}$

Solving we get

$v=\dfrac{f(2f+x)}{f+x}$

We have $u+v=1$

or

$2f+x+\dfrac{f(2f+x)}{f+x}=1$

or

$\dfrac{(2f+x)^2}{f+x}<1$

$(2f+x)^2<(f+x)$

This is valid only when $f<0.25m$

An object is placed at a distance $2 f$ from the pole of a convex mirror of focal length $f$. The linear magnification is:

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $\displaystyle \frac {1}{3}$

  2. $\displaystyle \frac {2}{3}$

  3. $\displaystyle \frac {3}{4}$

  4. 1

Show answer
Correct Option: A
Explanation:

$\displaystyle \frac {1}{V} - \frac {1}{2f} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac {3}{2f}  \Rightarrow v = \frac{2}{3}f$
$\therefore m = \displaystyle \frac {u}{v} = \frac{2}{3} \frac{f}{2f} = \frac {1}{3}$

The linear magnification for a mirror is the ratio of the size of the image to the size of the object, and is denoted by $'m'$. Then $m$ is equal to (symbols have their usual meanings)

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. True

  2. False

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Correct Option: B
Explanation:

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$.........(1)


multiplyng by u in eq.(1)

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

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

$\dfrac{u-f}{f}=\dfrac{u}{v}$

$\dfrac{f}{u-f}=\dfrac{v}{u}$

as $m=\dfrac{v}{u}$

hence, $m=\dfrac{f}{u-f}$

If linear magnification for a spherical mirror is $\dfrac{3}{2}$, then we may write: (symbols have their usual meanings) 
physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $f=\dfrac{u}{2}$

  2. $f=\dfrac{3u}{2}$

  3. $f=\dfrac{3u}{5}$

  4. None of these

Show answer
Correct Option: C
Explanation:

Mirror equation is: $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Multiplying both sides by $u$, we get:
$\dfrac{u}{f}=\dfrac{u}{v}+1$
Or  $\dfrac{u}{v}=\dfrac{u}{f}-1=\dfrac{u-f}{f}$
Or  $\dfrac{v}{u}=\dfrac{f}{u-f}$
Now, magnification, $m=\dfrac{v}{u}=\dfrac{3}{2}$
$\therefore$ $\dfrac{f}{u-f}=\dfrac{3}{2}$
Solving the above equation we get  $\dfrac{5}{2}f=\dfrac{3}{2}u$
 $\implies f=\dfrac{3u}{5}$

Magnification produced by a convex mirror is always:

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. equal to 1

  2. less than 1

  3. more than 1

  4. zero

Show answer
Correct Option: B
Explanation:

A convex mirror always creates a virtual image which is diminished. So, magnification produced by convex mirror is always less than one.

If magnification is positive, the nature of the image is:

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. real and inverted

  2. virtual and erect

  3. real

  4. none of these

Show answer
Correct Option: B
Explanation:

$\text{Magnification}=\dfrac{\text{Image  size}}{\text{Object  size}}$

According to new Cartesian sign convention, size of height of real and inverted image is considered negative and that of virtual and erect image is considered positive.
The heigh of the object, being erect is considered positive always.
So, for positive magnification, the ratio, mentioned above, will be positive. This implies that the image height will also be positive.
The image will be virtual and erect.

A concave mirror is made from a hollow sphere of radius of curvature 30 cm.  If an object of height 2 cm is placed at 10 cm from the pole of the mirror, determine the size of the image :

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. 3 cm

  2. 6 cm

  3. 12 cm

  4. 24 cm

Show answer
Correct Option: B
Explanation:

Given: $u=-10cm$


$f=-15cm$

$h _{o}=2cm$

To find, $h _{o}$

From mirror formula,

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

$\dfrac{1}{v}-\dfrac{1}{10}=-\dfrac{1}{15}$

$\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{15}=\dfrac{5}{150}$

$v=30cm$

$magnification=\dfrac{h _{i}}{h _{o}}=\dfrac{-v}{u}$

$\dfrac{h _{i}}{2}=\dfrac{-30}{-10}$

$h _{i}=6cm$

An object is placed on the principal axis of a concave mirror at a distance of 60 cm.  If the focal length of the concave mirror is 40 cm then determine the magnification of the obtained image.

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. 4

  2. -2

  3. -4

  4. +2

Show answer
Correct Option: B
Explanation:

Given: $u=-60 cm$     $f=-40 cm$
To find : $m$
Solution: From mirror formula
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{v}-\dfrac{1}{60}=\dfrac{-1}{40}$
$\dfrac{1}{v}=\dfrac{1}{60}-\dfrac{1}{40}$
$v=-120 cm$
Hence magnification is given by
$m=\dfrac{-v}{u}=-\dfrac{(-120)}{(-60)}$
$m=-2$

Calculate the magnification of an object if it is kept at a distance of $3 cm$ from a concave mirror of focal length $4 cm$:

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $3$

  2. $6$

  3. $9$

  4. $4$

Show answer
Correct Option: D
Explanation:

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

$\dfrac{1}{-3}+\dfrac{1}{v}=\dfrac{1}{-4}$
$\dfrac{1}{v}=\dfrac{1}{12}$
v= 12 cm
Magnification
$=\dfrac{12}{3}=4$

A concave mirror produces $2$ times magnified real image of an object placed at $5 cm$ in front of it. Where is image located?

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $10 cm$

  2. $-5 cm$

  3. $-10 cm$

  4. $2.5 cm$

Show answer
Correct Option: C
Explanation:

Magnification, $m = \dfrac {-v}{u}$
Given, magnification, $m = -2$

$-2 = \dfrac {-v}{-5 cm}$
Therefore, object distance, $v = -2\times 5 cm = -10 cm$