Tag: reflection of light at curved surfaces

Questions Related to reflection of light at curved surfaces

Ratio of the size of the image to the size of the object is known as:

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

  1. Focal plane

  2. Transformation ratio

  3. Efficiency

  4. None of these

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Correct Option: D
Explanation:

Ratio of the size of the image to the size of the object is known as magnification. It is given by $m = v/u$

In case of a real and inverted image, the magnification of a mirror 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. Positive

  2. Negative

  3. Zero

  4. Infinity

Show answer
Correct Option: B
Explanation:

We know,
Magnification(M)$=\dfrac{height  \ of\   image({h} _{i})}{height\   of  \ object({h} _{o})}$
Here,image is inverted so the $height \  of \  image({h} _{i})$will be negative.
Hence, the magnification of a mirror is negative.

Which of the following quantity does not have any unit?

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

  1. Velocity of light

  2. Light year

  3. Magnification

  4. Power of a lens

Show answer
Correct Option: C
Explanation:

Magnification does not have any unit as it is the ratio of same quantity.

The expression for the magnification of a spherical mirror in the terms of focal length (f) and the distance of the object from mirror (u) 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. $\frac{-f}{u-f}$

  2. $\frac{f}{u+f}$

  3. $\frac{-f}{u+f}$

  4. $\frac{f}{u-f}$

Show answer
Correct Option: D
Explanation:
Equation of spherical mirror is $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ , where v is the image distance.
solving,
Replacing $v$ with $v=mu$ , 
$\dfrac{1}{f} = \dfrac{1}{u}  (1 + \dfrac{1}{m} ) $
 $u = f (\dfrac{1}{m} +1)$ where m = magnification $= \dfrac{v}{u}$
$u =\dfrac{ f}{m} + f$
$m = \dfrac{f }{ (u - f)}$

A short linear object of length $L$ lies on the axis of a spherical mirror of focal length $f$ at a distance $u$ from the mirror. Its image has an axial length $L$ equal to :

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

  1. $L{ \left[ \cfrac { f }{ \left( u-f \right) } \right] }^{ 1/2 }$

  2. $L{ \left[ \cfrac { u+f }{ \left( f \right) } \right] }^{ 1/2 }$

  3. $L{ \left[ \cfrac { u+f }{ \left( f \right) } \right] }^{ 2 }$

  4. $L{ \left[ \cfrac { f }{ \left( u-f \right) } \right] }^{ 2 }$

Show answer
Correct Option: D
Explanation:

From mirror formula,       $\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } $


On differentiating, we get


      $\cfrac { -dv }{ { v }^{ 2 } } -\cfrac { du }{ { u }^{ 2 } } =0\\ \therefore dv=-du{ \left( \cfrac { v }{ u }  \right)  }^{ 2 }\\ as\quad \cfrac { v }{ u } =\cfrac { f }{ u-f } \\ \therefore dv=-du{ \left[ \cfrac { f }{ u-f }  \right]  }^{ 2 }\\ { L }^{ \prime  }=L{ \left[ \cfrac { f }{ u-f }  \right]  }^{ 2 }$

If an object is placed at a distance of 20cm from the pole of a concave mirror, the magnification of its real image is 3. If the object is moved away from the mirror by 10cm, then the magnification is -1.

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: A
Explanation:

$M= \frac{f}{f-d _0}$ and real image has M negative

$-3= \frac{f}{f-20}$

$-3f+60=f$

$f=15 cm$

$M= \frac{15}{15-30}$

$M= -1$

A convex lens is given, for which the minimum distance between an object and its rel image is $40cm$. An object is placed at a distance of $15cm$ from this lens. The liner magnification of adjustment will be 

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

  1. $\dfrac{5}{3}$

  2. $-2$

  3. $2$

  4. $\dfrac{1}{2}$

Show answer
Correct Option: A
Explanation:

Given object distance $u=15$ cm

Distance between object and real image produced $=40 $cm
Thus image distance $v=40-15=25$ cm
Also we know linear magnification,
$m=\dfrac{-v}{u}=\dfrac{-25}{-15}=\dfrac{5}{3}$ 

An object is placed in front of a concave mirror of radius of curvature 15 cm, at a distance of 10 cm, the position and nature of the image formed 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. $+30 cm, virtual \ and \ erect$

  2. $+30 cm, real \ and \ inverted$

  3. $-30 cm, virtual \ and \ erect$

  4. $-30 cm, real \ and \ inverted$

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Correct Option: D

An object of length $6\ cm$ is placed on the principle axis of a concave mirror of focal length $f$ at a distance of $4\ f$. The length of the image will be

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

  1. $2\ cm$

  2. $12\ cm$

  3. $4\ cm$

  4. $1.2\ cm$

Show answer
Correct Option: A
Explanation:

Given that,

The object distance $u=-6\,cm$

Now, magnification is

  $ m=\dfrac{I}{O} $

 $ m=\dfrac{f}{f-u} $

 $ \dfrac{I}{6}=\dfrac{-f}{-f-\left( -4f \right)} $

 $ I=-2\,cm $

Hence, the length of image is -$2\ cm$

An astronomical telescope has focal lengths $100$ & $10$cm of objective and eyepiece lens respectively when final image is formed at least distance of distinct vision,magnification power of telescope will be,

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

  1. -15

  2. -14

  3. -17

  4. -19

Show answer
Correct Option: B
Explanation:

Given focal length of eye piece${f} _{e}=10cm\$

focal length of objective${f} _{o}=100 cm\$
Also we know least distance $D=25 cm\$ 
Magnifying power $M=\dfrac{-{f} _{0}}{{f} _{e}}(1+\dfrac{{f} _{e}}{D})\$
$M=-\dfrac{100}{10}(1+\dfrac{10}{25})\$
$M=-14$