0

Focus and focal length - class-XII

Attempted 0/36 Correct 0 Score 0

The radii of curvature of the surfaces of a double convex lens are 20 cm and 40 cm respectively, and its focal length is 20 cm. What is the refractive index of the material of the lens.? 

  1. $\dfrac{5}{2}$

  2. $\dfrac{4}{3}$

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

  4. $\dfrac{4}{5}$


Correct Option: C
Explanation:

Here $R _1$ = 20 cm, $R _2$ = -40 cm, f = 20 cm
Using lens maker's formula we get,
$\dfrac{1}{20} \, = \, (\mu \, - \, 1) \left ( \dfrac{1}{20} \, + \, \dfrac{1}{40} \right )$
$\dfrac{1}{20} \, = \, (\mu \, - \, 1) \dfrac{3}{40} \, \Rightarrow \, \mu \, = \, \dfrac{5}{3}$

The focal length of a plane mirror is __________ .

  1. positive

  2. negative

  3. zero

  4. infinite


Correct Option: D
Explanation:

The radius of curvature of a plane mirror is infinite so focal length will also be infinite.

If we say that the focal length of a spherical mirror is $n$ times its radius of curvature, then $n$ must be

  1. $2.0$

  2. $1.5$

  3. $0.2$

  4. $0.5$


Correct Option: D
Explanation:

since we know that radius of curvature is 2 times focal length i.e R=2f

hence (f=0.5R)

The focal length of a spherical mirror is half of the radius of curvature

  1. For all rays

  2. Only for paraxial rays near the principal axis

  3. For those rays which are far from the principal axis

  4. For those rays which subtend extremely large angles with the axis


Correct Option: B
Explanation:
The rays that are near the principal axis (paraxial rays) and parallel to it converge to a single point on the axis after emerging from the spherical mirror. This point is called the focal point F of the lens.
And this is half of the radius of the curvature in spherical mirror.

State whether true or false.
The focal length of a spherical mirror is double its radius of curvature.

  1. True

  2. False


Correct Option: B
Explanation:

Focal length of  spherical mirror is half the radius of curvature i.e f $=\cfrac { R }{ 2 } $ .

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

  1. half

  2. twice

  3. thrice

  4. equal to


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?

  1. $100 cm$

  2. $25 cm$

  3. $50 cm$

  4. $400 cm$


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.

  1. $50 cm$

  2. $30 cm$

  3. $25 cm$

  4. $100 cm$


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:

  1. Plane mirror of small aperture

  2. Spherical mirrors of small aperture

  3. Plane mirrors of large aperture

  4. Spherical mirrors of large aperture


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.

  1. 24 cm

  2. 12 cm

  3. 6 cm

  4. 18 cm


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

  1. $+ 15.0 \mathrm { cm }$

  2. $+ 7.5 \mathrm { cm }$

  3. $- 7.5 \mathrm { cm }$

  4. $- 15.0 \mathrm { cm }$


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

  1. $6 cm$

  2. $4.5 cm$

  3. $9 cm$

  4. $4 cm$


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

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

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

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

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


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 $Km/h$?

  1. $3$

  2. $4$

  3. $5$

  4. $6$


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 ?

  1. 5 cm

  2. 7.5 cm

  3. -30 cm

  4. 12.5 cm


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.

What will be the height of image when an object of $2\ mm$ is placed at a distance $20 \ cm$ infront of the axis of a convex mirror of radius of curvature $40\ cm$ ?

  1. $20\ mm$

  2. $10\ mm$

  3. $6\ mm$

  4. $1\ mm$


Correct Option: D
Explanation:

$\begin{array}{l} \dfrac { 1 }{ V } +\dfrac { 1 }{ \mu  } =\dfrac { 1 }{ f }  \ \dfrac { 1 }{ V } =\dfrac { 1 }{ { 10 } } ,V=10cm \end{array}$

Height of object
$\begin{array}{l} =2mm=\dfrac { 1 }{ 5 } cm \ \dfrac { { Hi } }{ { Ho } } =\dfrac { V }{ 4 }  \ \dfrac { { Hi } }{ { 1/5 } } =\dfrac { { 10 } }{ { -20 } } \Rightarrow Hi=-\dfrac { 1 }{ { 10 } }  \ 1mm\, \, above\, \, axis \end{array}$

A convexo-concave diverging lens is made of glass of refractive index 1.5 and focal length 24 cm. Radius of curvature for one surface is doubled that of the other. Then radii of curvature for the two surface are (in cm) :

  1. 6 , 12

  2. 12, 24

  3. 3, 6

  4. 18, 36


Correct Option: D

An object is placed at 15 cm from a convex lens of focal length 10 cm . Where should another convex mirror of radius 12 cm placed such that image will coincide with object

  1. 18 cm

  2. 17 cm

  3. 14 cm

  4. 20 cm


Correct Option: A

Your are asked to design shaving mirror assuming that a person keep it 10 cm from his face and views the magnified image of the face at the closest comfortable distance of 25 cm. the reduice of curvature of the mirror would then be 

  1. 24 cm

  2. 60 cm

  3. -24 cm

  4. 30 cm


Correct Option: A

A spherical surface of radius of curvature $R$ separates air (refractive index 1.0) from glass (refractive index 1.5).The centre of curvature is in the glass. A point object $P$ placed in air is found to have a real image $Q$ in the glass. The line $PQ$ cuts the surface at a point $\mathbf { O } \text { and } \mathbf { P O }= \mathrm { OQ }$.Find the distance of object from the spherical surface.

  1. 3R

  2. 5R

  3. R

  4. 2R


Correct Option: A

A $10\mathrm { mm }$ long awl pin is placed vertically In front of a concave mirror. A $5\ mm$ long image of the awl pin is formed at $30\mathrm { cm }$ in front of the mirror, The focal length of this mirror is 

  1. $- 30 \mathrm { cm }$

  2. $- 20 \mathrm { cm }$

  3. $- 40 \mathrm { cm }$

  4. $- 60 \mathrm { cm }$


Correct Option: B

A point object is placed on principal axis of concave mirror of radius of curvature 10 cm at a distance 21 cm from pole of the mirror.  A glass slab of thickness 3 cm and refractive index 1.5 is placed between object and mirror $.$ Find the imaged position of the image formed. 

  1. 4

  2. 3

  3. 16.5

  4. 5


Correct Option: C

A converging bundle of light rays in the shape of cone with a vertex angle of 45 falls on a circular diaphragm of 20 cm diameter. A lens with power 5 D is fixed in the diaphragm. Diameter of face of lens is equal to that of diaphragm. If the vertex angle of new cone is 

  1. $
    \cfrac { 3 d } { 4 }
    $

  2. $
    \cfrac { 5 d } { 4 }
    $

  3. $
    2 d
    $

  4. $
    \cfrac { 3 } { 2 } d
    $


Correct Option: B

All the following statements are correct except

  1. The magnification produced by a convex mirror is always less than one.

  2. A virtual, crect , same-size image can be obtained by using a plane mirror.

  3. A virtual, crect , magnified image can be formed by using a convex mirror.

  4. A virtual, inverted , same-sized image can be formed by using a convex mirror.


Correct Option: A

If a spherical mirror is immersed in a liquid its focal length

  1. Decreases

  2. Remains same

  3. Increases

  4. Increases and Decreases


Correct Option: C

The suns diameter is $1.4\times { 10 }^{ 9 }m$ and its distance from the earth is ${ 10 }^{ 11 }m$. The diameter of its image, formed by a convex mirror of focal length 2m will

  1. 0.7 cm

  2. 1.4 cm

  3. 2.8 cm

  4. 10 cm


Correct Option: C

The distance of real object when a concave mirror produces a real image of magnification $'m'$ is ($f$ is focal length)

  1. $\left(\frac{m - 1}{m}\right) f$

  2. $\left(\frac{m + 1}{m}\right) f$

  3. $(m-1)f$

  4. $(m+1)f$


Correct Option: D

Converging rays are incident on a convex spherical mirror so that their extensions intersect  $30 cm$  behind the mirror on the optical axis. The reflected rays form a diverging beam, so that their extensions intersect the optical axis  $1.2 m$  from the mirror. The focal length of the mirror is

  1. $40{ cm }$

  2. $60{ cm }$

  3. $30{ cm }$

  4. $24{ cm }$


Correct Option: A

Magnification produced by a rear view mirror fitted in vehicles

  1. is less than one

  2. is more than one

  3. is equal to one

  4. can be more than or less than one depending upon the position of the object in front of it.


Correct Option: A

A concave mirror of radius of curvature 40 cm forms an image of an object placed on the principal axis at a distance 45 cm in front of it. Now if the system (including object) is completely immersed in water $(\mu=1.33)$, then:

  1. the image will shift towards the mirror.

  2. the magnification will reduce.

  3. the image will shift away from the mirror and magnification will increase.

  4. the position of the image and magnification will not change.


Correct Option: D
Explanation:

Mirror operates on the principle of Laws of reflection. 

Therefore, focal length does not depend upon the medium and object distance is also unchanged, so their will be no change in the image distance, correspondingly magnification will also remain unchanged.

The radius of curvature for a plane mirror is

  1. Positive

  2. Negative

  3. Infinite

  4. None of these


Correct Option: C
Explanation:

Plane mirrors are not curved.

The distance at which an object should be placed in front of a convex lens of focal length 10 cm to obtain a real image double the size of object will be:

  1. 30 cm

  2. 15 cm

  3. 5 cm

  4. 10 cm


Correct Option: B
Explanation:

Convex lens gives the real and double-sized image when the object is placed exactly between the focus and radius of curvature.
We have, $\displaystyle \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$m = \displaystyle \frac{v}{u} = 2$ or $v = 2u$


$\therefore \displaystyle \frac{1}{f} = \frac{1}{2u} - \frac{1}{-u} = \frac{1}{2u} + \frac{1}{u} = \frac{3}{2u}$

or $\displaystyle \frac{1}{10} = \frac{3}{2u}$ or $u = 15 cm$

The focal length of a convex mirror is $10cm$. Its radius of curvature will be:

  1. $5cm$

  2. $10cm$

  3. $20cm$

  4. $30cm$


Correct Option: C
Explanation:

The relation between focal length and Radius of curvature is,

$R=2f$
Here focal length is 10cm so,
$R=2\times 10 = 20cm$

An object is placed at the centre of curvature of a concave mirror of radius of curvature $20$cm. The nature and position of the image shall be.

  1. Virtual and $15$cm from the mirror

  2. Real and $20$cm from the mirror

  3. Virtual and $20$cm from the mirror

  4. Real and $10$cm from the mirror


Correct Option: B
Explanation:

Image of an object placed at center of curvature is inverted, real and of the same size and is formed at the center of curvature. Hence image will be real and at the center of curvature (20 cm from mirror).

Formula of focal length in convex lens is

  1. $\displaystyle f = \frac{u+v}{u-v}$

  2. $\displaystyle f = \frac{u\times v}{u-v}$

  3. $\displaystyle f = \frac{u-v}{u+v}$

  4. $\displaystyle f = \frac{u+v}{u+v}$


Correct Option: B

An object is placed at a distance of $50\ cm$ from a convex mirror. A plane mirror is placed in front of the convex mirror in such a way that it covers half of the convex mirror. If the distance between object and plane mirror is $30\ cm$ then there is no parallax between the images formed by two mirrors, the radius of curvature of convex mirror will be :

  1. $50\ cm$

  2. $25\ cm$

  3. $12.5\ cm$

  4. $100\ cm$


Correct Option: A
- Hide questions