Tag: physics

Questions Related to physics

The focal length of a concave mirror is 50 cm where an object is to be placed so that its image is two times magnified, real and inverted :

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. 75 cm

  2. 72 cm

  3. 63 cm

  4. 50 cm

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Correct Option: A
Explanation:

Given,

Focal length, $f=50\,cm$

Magnification, $m=2$

$ m=\dfrac{v}{u} = 2$

$ v=2u $

From mirror formula,

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

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

$ u=\dfrac{3f}{2}=\dfrac{3\times 50}{2}=75\ cm $

Object Is placed at $75\,cm$ from the mirror.

A convex mirror has a focal length $f$.A real object is placed at a distance $f$ in front of it from the pole, produces an image at:

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. $\infty$

  2. $f$

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

  4. $2f$

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Correct Option: C

The distance between an object its doubly magnified by a concave mirror of focal length $f$ is

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. $3 f/2$

  2. $2 f/3$

  3. $3\ f$

  4. Depend on whether the image is real or virtual

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Correct Option: A
Explanation:

Given that,

Focal length $=f$

We know that, the magnification is

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

Now, magnification is double and image is real

  $ -2=\dfrac{-v}{u} $

 $ v=2u $

Now, using formula of mirror

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

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

 $ \dfrac{1}{f}=\dfrac{3}{2u} $

 $ u=\dfrac{3f}{2} $

Hence, the distance is $\dfrac{3f}{2}$ 

An object $2.5\ cm$ high is placed at a distance of $10\ cm$ from a concave mirror of radius of curvature $30\ cm$. The size of the image is:

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. $9.2\ cm$

  2. $10.5\ cm$

  3. $5.6\ cm$

  4. $7.5\ cm$

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Correct Option: D
Explanation:

Given that,

Distance $u=-10\,cm$

Height $h=2.5\,cm$

Radius of curvature $R=30\,cm$

We know that,

  $ -f=\dfrac{R}{2} $

 $ -f=\dfrac{30}{2} $

 $ f=-15\,cm $

Now, using mirror’s formula

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

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

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

 $ v=30\,cm $

Now, the magnification is

  $ m=\dfrac{-v}{u} $

 $ m=\dfrac{30}{10} $

 $ m=3 $

We know that,

  $ m=\dfrac{h'}{h} $

 $ 3=\dfrac{h'}{2.5} $

 $ h'=7.5\,cm $

Hence, the size of the image is $7.5\ cm$

The focal length $f$ of a mirror is given by $\cfrac{1}{f}=\cfrac{1}{u}+\cfrac{1}{v}$, where $u$ and $v$ represent object and image distances, respectively

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}$

  2. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{v}+\cfrac{\Delta v}{u}$

  3. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}-\cfrac{\Delta (u+v)}{u+v}$

  4. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}+\cfrac{\Delta U}{u+v}+\cfrac{\Delta v}{u+v}$

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Correct Option: D
Explanation:

The focal length $f$ of a mirror is given by
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$. . . . . . . .(1)
where $v=$ image distance
$u=$ object distance
from equation (1),
$f=\dfrac{uv}{u+v}=uv(u+v)^{-1}$
Taking log both sides we get
$logf=logu+log v+log(u+v)^{-1}$
$log f=logu+logv-log(u+v)$
Differentiating with respect to each variable
$\dfrac{\Delta f}{f}=\dfrac{\Delta u}{u}+\dfrac{\Delta v}{v}+\dfrac{\Delta (u+v)}{u+v}$
$\dfrac{\Delta f}{f}=\dfrac{\Delta u}{u}+\dfrac{\Delta v}{v}+\dfrac{\Delta u}{u+v}+\dfrac{\Delta v}{u+v}$
The correct option is D.

A concave lens of focal length $f$ produces an image $(1/x)$ of the size of the object. The distance of the object from the lens is

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. $(x - 1)f$

  2. $(x + 1)f$

  3. $\dfrac{(x - 1)}{x}$ $f$

  4. $\dfrac{(x + 1)}{x}$ $f$

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Correct Option: B

A concave spherical mirror, forms a 40 cm high real image of an object, whose height is 10 cm. The radius of the mirror is 60 cm. Find the distance from the object to its image.

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. 97.5 cm

  2. 75 cm

  3. 90cm

  4. 177.5 cm

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Correct Option: D

When a ray of light parallel to the principle axis is incident on a concave mirror$,$ the reflected ray

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. Passes through C

  2. Passes through F

  3. Passes midway between P and F

  4. retraces its path

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Correct Option: B

A mirror faces the negative x-axis. (Normal to its reflecting surface is$- \hat { i } )$  While a particle starts moving such that its image is formed in the mirror. At a certain instant the velocity of the particles is $3 \hat { i } + 4 \hat { j } + 5 \hat { k }$ and that of the mirror is $\hat { 1 } - \hat { j } + \hat { k }$ Choose the correct options.

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. Magnitude of relative velocity of the image w.r.t mirrror is$\sqrt { 45 }$

  2. Magnitude of relative velocity of image w.r.t object is 4

  3. Magnitude of relative velocity of image w.r.t mirrror is $\sqrt { 45 }$

  4. Absolute velocity of the image w.r.t ground is$\sqrt { 42 }$

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Correct Option: A,C
Explanation:

$\begin{array}{l} In\, y-direction \ { V _{ o } }={ V _{ i } } \ \Rightarrow { V _{ i } }=4\widehat { j }  \ In\, z-direction \ { V _{ i } }={ V _{ o } } \ \Rightarrow { V _{ i } }=5\widehat { k }  \ In\, x-direction \ { V _{ IM } }=-{ V _{ oM } } \ \Rightarrow { V _{ IM } }=-{ V _{ oM } } \ \Rightarrow { V _{ I } }-{ V _{ M } }={ V _{ M } }-{ V _{ o } } \ \Rightarrow { V _{ I } }=2{ V _{ M } }-{ V _{ o } } \ =2\left[ { \widehat { i }  } \right] -3\left[ { \widehat { i }  } \right]  \ =-\widehat { i }  \ \therefore \overrightarrow { { V _{ I } } } =-\widehat { i } +4\widehat { j } +5\widehat { k }  \ \therefore \overrightarrow { { V _{ iM } } } =-2\widehat { i } +5\widehat { j } +4\widehat { k }  \ \therefore \left| { \overrightarrow { { V _{ iM } } }  } \right| =\sqrt { 45 }  \end{array}$

Hence,
option $(A)$ and $(C)$ are both correct answer.

A thin. rod of length f/ 3 is placed along the principal axis of a concave mirror of focal length f such that its image which is real and elongated, just touches one end of the rod. What is its magnification ?

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. +2

  2. -3

  3. -1.5

  4. -2

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Correct Option: C