Reflection From Spherical Mirrors

A spherical mirror, as the name suggests, has the shape of a section of a hollow sphere.

A spherical mirror is a mirror whose reflecting surface is made by the part of a hollow sphere. 

Suppose a hollow sphere has a polished mirror surface on the inside as well outside. 

By removing a section of the sphere, a double-sided spherical mirror is obtained with a concave reflecting surface on one side and a convex reflecting surface on the other.

1.0Concave Mirror

A spherical mirror in which the reflection of light takes place at bent-in surface is called ‘concave mirror’.

Concave mirror is also called ‘converging mirror’. This is because the parallel beam of light after reflection, converge at a single point.

The concave reflecting surface is curved inwards. The convex reflecting surface is curved outwards.

2.0Convex Mirror

A spherical mirror in which the reflection of light takes place at bulging-out surface is called ‘convex mirror’. 

Convex mirror is also called ‘diverging mirror’. This is because the parallel beam of light after reflection appears to diverge from a single point.

Some basic terms related to spherical mirrors

Centre of curvature : The point in space that represents the centre of the hollow sphere from which the spherical mirror was cut is called ‘centre of curvature’.

The centre of the hollow sphere from which the spherical mirror is formed is called ‘centre of curvature’.  

Pole (or vertex) : The middle point on the surface of a spherical mirror is called ‘pole’.

The geometric centre of the curved mirror surface is called ‘pole’.

Radius of curvature : The radius of hollow sphere from which the mirror is formed is called ‘radius of curvature’.

The distance between the centre of curvature and the pole of a spherical mirror is called ‘radius of curvature’.

Principal axis : A line passing through the pole and the centre of curvature of the spherical mirror is called ‘principal axis’.

An imaginary line drawn through the pole, perpendicular to the surface of the spherical mirror at the pole is called ‘principal axis’.

Principal focus : The point on the principal axis where all rays parallel to principal axis, either converge or appear to diverge after reflection is called ‘principal focus’.

For concave mirrors, centre of curvature and principal focus are real. For convex mirrors, centre of curvature and principal focus are  virtual.

Focal length : The distance between the focus and pole of a spherical mirror is called ‘focal length’.

Focal plane : A plane passing through focus and perpendicular to principal axis is called ‘focal plane’.

Aperture : The diameter of the circular cross-section of the spherical mirror is called ‘aperture’ (AB). It represents the size of the mirror. More the aperture of a spherical mirror, more will be its size. Thus, it will collect more light forming brighter images after reflection.

3.0Rules to Obtain Images in Spherical Mirrors

Concave mirrors

(1) The ray parallel to the principal axis, after reflection, passes through the principal focus F of a concave mirror.

(2) A ray passing through the principal focus in a concave mirror, is reflected parallel to the principal axis.

(3) A ray passing through the centre of curvature in a concave mirror, is reflected back along its own path .

Convex mirrors

(1) The ray parallel to the principal axis, after reflection, appears to diverge from the principal focus of a convex mirror.

(2) A ray which is directed towards the principal focus in a convex mirror, is reflected parallel to the principal axis.

(3) A ray directed towards the centre of curvature in a convex mirror, is reflected back along its own path.

Obtaining image of the sun through a concave mirror

1. Take a concave mirror and allow the sun rays to fall on it. Take paper and move it towards the concave mirror till you obtain a bright sharp spot of light on it. The spot obtained is the image of the sun. Now, measure the distance between paper and the concave mirror. This distance is an approximate focal length of the concave mirror. 

2. If this spot is kept on the paper for few minutes, the paper will start burning. This is because the light energy converts to heat energy.

Important : Avoid looking at the Sun directly or its image formed by the concave mirror as the intensity of sunlight may damage the eye.

4.0Image Formation by Concave Mirror

5.0Uses of Concave Mirrors

(1) Concave mirrors are used as shaving mirrors to see a larger image of the face.

(2) Concave mirrors are used as reflectors in car head lights, search lights, hand torches, table lamps, etc. to get powerful parallel beams of light.

(3) Concave mirrors are used in solar power plants to produce electricity.

(4) Concave mirrors are used by doctors to concentrate light on body parts like ears and eyes. Concave mirror can be used as a magnifier to see a larger image of an object when it is placed between pole and focus. 

(5) Concave mirrors are also used by dentists to see large images of the teeth of patients.

6.0Image formation by a Convex Mirror

The image formed by a convex mirror is always behind the mirror that is, it is always virtual and erect. Also, the size of image is always diminished, that is, its size is always smaller than that of the object .

The rays parallel to principal axis, after reflection, appears to diverge from the principal focus of the convex mirror [see figure (b)]. The image formed at the focus, behind the mirror is highly diminished. The image is virtual and erect.

Image formation by convex mirror

Uses of Convex Mirrors

Rear view mirrors : Convex mirrors are commonly used as rear-view (wing) mirrors in vehicles. These mirrors are fitted on the sides of the vehicle, enabling the driver to see traffic behind him/her to facilitate safe driving.

Convex mirror

Convex mirrors are preferred as rear view mirrors because they always give an erect, though diminished image. Also, they have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view much larger area than would be possible with a plane mirror [see figure (a) and figure (b)].

Street lamps :  Street lamps also use convex mirrors to diverge light over an extended area. 

You can see a full-length image of a tall building/tree in a small convex mirror. 

Sign convention for reflection by spherical mirrors

While dealing with the reflection of light by spherical mirrors, we follow a set of sign conventions called the new cartesian sign convention. In this convention, the pole (P) of the mirror is taken as the origin. The principal axis of the mirror is taken as the x-axis (X’X) of the coordinate system.

The conventions are as follows [see figure (a) and (b)] :

(1) The object is always placed to the left of the mirror. This implies that the light from the object falls on the mirror from the left-hand side.

(2) All distances parallel to the principal axis are measured from the pole of the mirror. All the distances along XX’ axis are measured from P.

(3) All the distances measured to the right of the origin (along + x-axis) are taken as positive while those measured to the left of the origin (along – x-axis) are taken as negative.

(4) Distances measured perpendicular to and above the principal axis (along + y-axis) are taken as positive. Distances measured perpendicular to and below the principal axis (along – y-axis) are taken as negative.

Distances along the direction of incident light are considered ‘positive’. Distances measured opposite to the direction of incident light are considered ‘negative’.

If image is virtual and erect i.e., above principal axis, its height is taken ‘positive’. If image is real and inverted i.e., below principal axis, its height is taken ‘negative’.

7.0Formulae Related to Spherical Mirrors

Relationship Between Radius of Curvature and Focus

The focal length of a spherical mirror is equal to half of its radius of curvature.

$f=\frac{R}{2}$

Mirror Formula

In a spherical mirror, the distance of the object from its pole is called the object distance (u). The distance of the image from the pole of the mirror is called the image distance (v). 

The relationship between object distance (u), the image distance (v) and the focal length (f) is given by mirror formula which is as given below,

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

Magnification (m) 

The ratio of height of image (h₂) to the height of object (h₁) is called ‘magnification’ or ‘linear magnification’.

$m=\frac{h_2}{h_1}$

The magnification (m) is also related to the object distance (u) and image distance (v). It can be expressed as:

$m=\frac{h_2}{h_1}=\frac{-v}{u}$

Also, magnification can be further expressed as,

$m=\frac{f}{f-u}=\frac{f-v}{f}$

A plane mirror forms a virtual, erect and same size image as that of the object thus, the magnification of a plane mirror is +1.

Some important points related to spherical mirrors

(a) Concave mirror

(1) Object distance, u = always negative.

(2) Image distance, v = positive, when an object is placed between P & F (virtual and erect image). v = negative, all other possible cases (real and inverted image).

(3) f = negative, R = negative

(4) For concave mirror, ‘m’ can be positive as well as negative. Also, |m| can be less than, equal to or greater than one.

(b) Convex mirror 

(1) Object distance, u = always negative. 

(2) Image distance, v = always positive (virtual and erect).

(3) f = positive, R = positive.

(4) Image is always diminished.

(5) For convex mirror, ‘m’ is always positive and |m| is always less than one. This is because it always forms a virtual, erect and diminished image of the object.

Solved Examples

1. A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image.

Solution

Given, radius of curvature, R = + 3 m; object distance, u = – 5 m ;

image distance, v = ? ; magnification, m = ? ;  Now, focal length, f = R/2 = +3/2 m

Mirror formula,  

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

$\frac{1}{v}+\frac{1}{-5}=\frac{1}{(+3/2)}$

$\frac{1}{v}-\frac{1}{5}=\frac{2}{3}$

$\frac{1}{v}=\frac{1}{5}+\frac{2}{3}=\frac{3+10}{15}=\frac{13}{15}$

$v=\frac{15}{13}=+1.15\text{m}$

Thus, the image is 1.15 m at the back of the mirror.

Now, magnification, 

$m=\frac{h_2}{h_1}=-\frac{v}{u}=-\frac{\left(+\frac{15}{13}\right)}{-5}=\frac{3}{13}=+0.23$

The image is virtual, erect and smaller in size by a factor of 0.23.

2. An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image ? Find the nature and the size of the image.

Solution

Given, object size, h₁ = + 4 cm ; object distance, u = – 25 cm ;

Focal length, f = –15 cm ; image distance, v = ? ; image size, h₂ = ?

Mirror formula,

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

$\frac{1}{v}+\frac{1}{(-25)}=\frac{1}{-15}$

$\frac{1}{v}-\frac{1}{25}=-\frac{1}{15}$    

$\frac{1}{v}=\frac{1}{25}-\frac{1}{15}=\frac{3-5}{75}=\frac{-2}{75}$

$v=\frac{-75}{2}=-37.5\text{cm}$

Now, magnification,

$m=\frac{h_2}{h_1}=-\frac{v}{u}$

$h_2=-\frac{v}{u}\times h_1=-\left(\frac{-75/2}{-25}\right)\times (4)=-\frac{3}{2}\times 4=-6\text{cm}$

The image is real, inverted and enlarged.

3. An object is placed at 10 cm in front of a concave mirror of radius of curvature 15 cm. Find the magnification of the image.

Solution

Given, object distance, u = –10 cm ; radius of curvature, R = – 15 cm ;

image distance, v = ? ; magnification, m = ?

Focal length, f = –15/2 cm

Mirror formula,          

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

$\frac{1}{v}+\frac{1}{(-10)}=\frac{1}{(-15/2)}$       

$\frac{1}{v}-\frac{1}{10}=-\frac{2}{15}$                 

$\frac{1}{v}=\frac{1}{10}-\frac{2}{15}=\frac{3}{30}-\frac{4}{30}=-\frac{1}{30}$

$v=-30\text{cm}$

Now magnification, 

$m=-\frac{v}{u}=-\frac{(-30)}{(-10)}=-3$

4. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved ?

Solution

Given, object distance, u = – 27 cm ; radius of curvature, R = – 36 cm ; object size, h₁ = +2.5 cm ; image distance, v = ? ; image size, h₂ = ?

Now, focal length, 

f = R/2 = –36/2 cm = – 18 cm

Mirror formula,        

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

$\frac{1}{v}+\frac{1}{(-27)}=\frac{1}{(-18)}$

$\frac{1}{v}-\frac{1}{27}=-\frac{1}{18}$

$\frac{1}{v}=\frac{1}{27}-\frac{1}{18}=\frac{2}{54}-\frac{3}{54}=-\frac{1}{54}$

$v=-54\text{cm}$

Now magnification, 

$h_2=-\frac{v}{u}\times h_1=-\frac{(-54)}{(-27)}\times (+2.5)=-5\text{cm}$

The image is real, inverted and enlarged.

When the candle is moved closer to the mirror, the screen has to be moved away from the mirror. But, when candle is at a distance less than 18 cm (i.e., less than its focal length) from the mirror, image formed will be virtual which is not possible to obtain on the screen.

5. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Solution

Given, object distance, u = – 12 cm ; focal length, f = + 15 cm ; object size, h₁ = +4.5 cm ; image distance, v = ? ; magnification, m = ?, image size, h₂ = ? 

Mirror formula,

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

$\frac{1}{v}+\frac{1}{(-12)}=\frac{1}{(+15)}$

$\frac{1}{v}-\frac{1}{12}=\frac{1}{15}$

$\frac{1}{v}=\frac{1}{15}+\frac{1}{12}=\frac{4}{60}+\frac{5}{60}=\frac{9}{60}$

$v=\frac{60}{9}\text{cm}=+6.67\text{cm}$

Now, magnification, 

$m=-\frac{v}{u}=-\frac{(+60/9)}{(-12)}=+\frac{5}{9}=+0.55$

Also, magnification

$m=\frac{h_2}{h_1}$

$+\frac{5}{9}=\frac{h_2}{(+4.5)}$

$h_2=+\frac{5}{9}\times (+4.5)=+2.5\text{cm}$

Image is virtual, erect and diminished.

If the needle (object) is moved farther from the mirror, its image moves away from the mirror i.e., from pole towards the focus. The image remains virtual and eret but it gradually decreases in size. When the object becomes infinitely far away, the image is formed at the focus and it is a point sized image. But, the image never goes beyond the focus in a convex mirror.

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