Tag: physics

Questions Related to physics

The focal length of a spherical mirror is ____ its radius of curvature.

physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

  1. half

  2. twice

  3. thrice

  4. equal to

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Correct Option: A
Explanation:

Focal length of  spherical mirror is half the radius of curvature i.e $f =\dfrac R2$

Focal length of a spherical mirror is $200 cm$. What will be its radius of curvature?

physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

  1. $100 cm$

  2. $25 cm$

  3. $50 cm$

  4. $400 cm$

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Correct Option: D
Explanation:
We know,
$ Focal\ length\ (f) = \dfrac{Radius\ of\ curvature\ (R)}{2} $

$ \Rightarrow Radius\ of\ curvature\ (R) = 2 \times Focal\ length\ (f) $

Given, 
Focal Length, $ f = 200\ cm $
$ \Rightarrow R = 2 \times f = 2 \times 200 = 400\ cm $
$ \Rightarrow Radius\ of\ curvature\ (R) = 400\ cm $

Hence, the correct answer is OPTION D.

A spherical mirror has radius of curvature equal to $50 cm$. Find the value of focal length.

physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

  1. $50 cm$

  2. $30 cm$

  3. $25 cm$

  4. $100 cm$

Show answer
Correct Option: C
Explanation:
We know,
$ Focal\ length\ (f) = \dfrac{Radius\ of\ curvature\ (R)}{2} $

Given, 
Radius of curvature, $ R = 50\ cm $
$ \Rightarrow f = \dfrac{R}{2} = \dfrac{50}{2} = 25\ cm $
$ \Rightarrow  Focal\ length\ (f) = 25\ cm $

Hence, the correct answer is OPTION C.

Radius of curvature is found to be equal to twice the focal length for:

physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

  1. Plane mirror of small aperture

  2. Spherical mirrors of small aperture

  3. Plane mirrors of large aperture

  4. Spherical mirrors of large aperture

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Correct Option: B
Explanation:

For spherical mirrors of small apertures, the radius of curvature is found to be equal to twice the focal length. We put this as $R = 2f$. This implies that the principal focus of a spherical mirror lies midway between the pole and centre of curvature.

A concave mirror is made by cutting a portion of a hollow glass sphere of radius $24$ cm. Find the focal length of the mirror.

physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

  1. 24 cm

  2. 12 cm

  3. 6 cm

  4. 18 cm

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Correct Option: B
Explanation:

The radius of curvature of the mirror $=24 cm.$
Thus the focal length
$=24 cm/2$
$=12 cm.$

An object is at a distance of  $10cm$  from a concave mirror and the image of the object is at a distance of  $30\mathrm { m }$ from the mirror on the same side as that of the object. The radius of curvature of the concave mirror is

physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

  1. $+ 15.0 \mathrm { cm }$

  2. $+ 7.5 \mathrm { cm }$

  3. $- 7.5 \mathrm { cm }$

  4. $- 15.0 \mathrm { cm }$

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Correct Option: C

A dobleconves lens of focal length $6 cm$ is made of glass of refractive index $1.5$ the radius of curvature of of one surface is double that of other surface. The value of small radius of curvature is

physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

  1. $6 cm$

  2. $4.5 cm$

  3. $9 cm$

  4. $4 cm$

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Correct Option: B

A real image of half the size is obtained in a concave spherical mirror with a radius of curvature of $40 cm$, the distance of object and its image will be

physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

  1. $30 cm\quad and \quad 60cm$

  2. $60 cm\quad and \quad 30cm$

  3. $15 cm\quad and \quad 30cm$

  4. $30 cm\quad and \quad 15cm$

Show answer
Correct Option: B
Explanation:
Lets, $u=$ object distance
$v=$ image distance
Given, 
$R=40cm$
$f=\dfrac{R}{2}=20cm$
magnification, $m=\dfrac{h}{2h}=\dfrac{v}{u}$ (for real image)
$v=\dfrac{u}{2}$. . . . (1)
By mirror formula,
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
$\dfrac{1}{20}=\dfrac{2}{u}+\dfrac{1}{u}$
$u=60cm$
From equation (1),
$v=\dfrac{60}{2}=30cm$
The correct option is B.

image of an object approching a convex mirror of radius of curvature 20 m along its optical axis so is observed to move from $\dfrac{25}{3}$ m to $\dfrac{50}{7}$m into 30s. what is the speed of the object in $Km/h$?

physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

  1. $3$

  2. $4$

  3. $5$

  4. $6$

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Correct Option: A
Explanation:

$A=20m$  $f=10m.$

From the mirror equation$:$
$\dfrac{1}{{{v _1}}} + \dfrac{1}{{{u _1}}} = \dfrac{1}{f};$
$\frac{1}{{25/3}} + \dfrac{1}{{{u _1}}} = \dfrac{1}{{10}};$
$ = {u _1} =  - 50\,m.$
furthermore$,$ when the picture of the question is at $50/7m.$
$\dfrac{1}{{{V _2}}} + \dfrac{1}{{{u _2}}} = \dfrac{1}{f}$
$\dfrac{1}{{50/7}} + \dfrac{1}{{{u _2}}} = \dfrac{1}{{10}}$
$ = {u _2} = 25m$
contrast out there of the protest$=50-25=25m$
speed$=$ relocation/time
$=25/30$
$5/6 m/sec$
speed in $km/h$ $ = 5/6 \times 18/5$
$ = 3\,km/h.$
Hence,
option $(A)$ is correct answer.

A convex mirror of radius of curvature 20 cm forms an image which is half the size of the object.How far is the object from the mirror ?

physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

  1. 5 cm

  2. 7.5 cm

  3. -30 cm

  4. 12.5 cm

Show answer
Correct Option: C
Explanation:

Radius of curvature $=20 cm$

So$,$ focal length $=10 cm$
We are given$,$ ${h _o}/2 = {h _i}$
$so,\,{h _o} = 2{h _i}$
$so,\,m = {h _i} = {h _o}$
$m = {h _o}/2{h _o}$
$so,\,m =  - v/u$
$1/2 =  - v/u$
$so,\,u =  - 2v$
By mirror formula$,$ 
$1/f = 1/v + 1/u$
$1/10 = 1/v - \left( { - 1/2v} \right)$
$1/10 = 1/v + 1/2v$
$so,\,1/10 = 2 + 1/2v$
$so,\,1/10 = 3/2v$
$so,\,2v = 30$
$so,\,v = 15\,cm$
$u =  - 2\left( v \right) =  - 30$
Hence,
option $(C)$ is correct answer.