Tag: reflection of light at curved surfaces

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Questions Related to reflection of light at curved surfaces

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

In case of a real and inverted image, the magnification of a mirror is:

  1. Positive

  2. Negative

  3. Zero

  4. Infinity
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $\frac{-f}{u-f}$
  2. $\frac{f}{u+f}$
  3. $\frac{-f}{u+f}$
  4. $\frac{f}{u-f}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $= \dfrac{v}{u}$
$u =\dfrac{ f}{m} + f$
$m = \dfrac{f }{ (u - f)}$
Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

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 :

  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. $L{ \left[ \cfrac { f }{ \left( u-f \right) } \right] }^{ 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


On differentiating, we get


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

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

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.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

$-3f+60=f$

$f=15 cm$

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

$M= -1$

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

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 

  1. $\dfrac{5}{3}$
  2. $-2$
  3. $2$
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,
$m=\dfrac{-v}{u}=\dfrac{-25}{-15}=\dfrac{5}{3}$ 

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

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 :

  1. $+30 cm, virtual \ and \ erect$
  2. $+30 cm, real \ and \ inverted$
  3. $-30 cm, virtual \ and \ erect$
  4. $-30 cm, real \ and \ inverted$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

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

  1. $2\ cm$
  2. $12\ cm$
  3. $4\ cm$
  4. $1.2\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 -$2\ cm$

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

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,

  1. -15

  2. -14

  3. -17

  4. -19
Reveal answer Fill a bubble to check yourself
B Correct answer
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})\$
$M=-14$