Tag: derivation of formula for curved mirrors

Questions Related to derivation of formula for curved mirrors

What is known as linear magnification of spherical mirrors?

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

  1. Ratio of size of image to size of object

  2. Ratio of shape of image to size of object

  3. Ratio of size of image to shape of object

  4. None

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Correct Option: A
Explanation:


Ratio of size of image to size of object is known as linear magnification of spherical mirrors.

Linear magnification (m) is the ratio of height of image to that of the object.

Magnification $(m) =$ ______

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

  1. $\dfrac {v}{u}$

  2. $\dfrac {u}{v}$

  3. $\dfrac {h _{o}}{h _{i}}$

  4. $\dfrac {h _{i}}{h _{o}}$

Show answer
Correct Option: D
Explanation:

Magnification is ratio of the size of the image $h _i$ to the size of the object $h _o$.

$m=\dfrac{h _i}{h _o} = \dfrac{-v}{u}$ where $u$ is object distance and $v$ is image distance.

For magnification in spherical mirrors object height 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. Negative.

  2. Positive.

  3. For real images positive.

  4. For virtual images negative.

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Correct Option: B
Explanation:

Magnification is the increase in the image size produced by spherical mirrors with respect to the object size. It is the ratio of the height of the image to the height of the object. The height of the object is always positive as the object is always above the principal axis. 

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

Show answer
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,
$m=\dfrac{-v}{u}=\dfrac{-25}{-15}=\dfrac{5}{3}$