Tag: reflection of light by curved surfaces

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Questions Related to reflection of light by curved surfaces

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

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. $- 15.0 \mathrm { cm }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

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. $4 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Lens maker's formula: 1/f = (n-1)(1/R1 - 1/R2). Given f=6, n=1.5, R1=x, R2=-2x (double-convex). 1/6 = (0.5)(1/x + 1/2x) = 0.5(3/2x) = 3/4x. So 4x = 18, x = 4.5 cm.

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

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. $30 cm\quad and \quad 15cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

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. $6$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $1\ mm$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$\begin{array}{l} =2mm=\dfrac { 1 }{ 5 } cm \ \dfrac { { Hi } }{ { Ho } } =\dfrac { V }{ 4 }  \ \dfrac { { Hi } }{ { 1/5 } } =\dfrac { { 10 } }{ { -20 } } \Rightarrow Hi=-\dfrac { 1 }{ { 10 } }  \ 1mm\, \, above\, \, axis \end{array}$

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First, find the image position from the lens: 1/v - 1/-15 = 1/10 => 1/v = 1/10 - 1/15 = 1/30. v = 30 cm. For the mirror to make the image coincide, the light must strike the mirror normally, meaning the image must be at the center of curvature. Mirror R = 12, so C = 12. Distance = 30 - 12 = 18 cm.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer