Light-reflection And Refraction

Concave mirrors are used as shaving mirrors to see a large image of the face and make a good shave. When the face is held within the focus of a concave mirror (not on the focus) then the enlarged image can be seen in the concave mirror.

we know,

If the height of the object is positive and the height of image is negative the magnification will be negative. So, negative sign in value of magnification indicates that image is real and inverted because a real image has always negative height i.e. real image is formed below principal axis.

For seeing an enlarged image I will use a concave mirror. When the face (AB) is held within the focus of a concave mirror then the enlarged image (A’B’) can be seen in the concave mirror. Note: If the face is at focus then enlarged image will be formed at infinity.

Refractive index of glass: 1.52 to 1.9

Refractive index of water: 1.33

Refractive index of diamond: 2.42

A substance having higher refractive index is optically denser then the substance of lower refractive index and light will travel slowest in the most optically dense substance i.e. having highest refractive index.

(a) Light travels slowest in the diamond having highest optical density.

(b) Light travels fastest in the water having lowest optical density.

Optical density of a substance is the degree to which it will slow down the ray of light passing through it.

Higher the refractive index of a substance, higher will be its optical density.

Optical density in ascending order will be :

Ice - 1.31

Crown glass - 1.52

Rock salt - 1.54

Dense flint glass - 1.65

If the Object is placed at distance 2f from a convex lens then a real and inverted image of same size as object will be formed. So, size of image formed = 1 cm

If the image formed by a lens is diminished and erect then then it must be a concave lens. Object(AB) placed anywhere on the principal axis, it will always form virtual, erect and diminished image(A’B’).

Given,

Focal length of concave lens = -50 cm

= m

= -0.50 m

To find: Power of lens?

We know,

Power of lens = -2 Dioptres.

Since the focal length of a concave lens is always negative so power of the concave lens is -2 diopters.

Two laws of reflection of light are:

First law: The incident ray, the reflected ray and the normal at the point of incidence, all lie in same plane. In the given figure ray OA, OB and ON lie in same plane.

Second law: The angle of reflection (r) is always equal to the angle of incidence (i).

Given,

Magnification by plane mirror= 1

We know,

Substituting value of magnification in this formula:

Height of object = Height of image

So, the given information signifies that image formed by a plane mirror is of same size as that of object.

Also,

Substituting value of magnification in this formula:

Object distance = - Image distance

Or image distance = - Object distance

So, the given information also tells that image formed by a plane mirror is at same distance behind the mirror as the object is in front of the mirror i.e. virtual and erect image.

Diagram 1

When a ray of light passes parallel to the principal axis of a concave lens, then it appears to be coming from its focus (F) after refraction through the lens.

Diagram 2:

When a ray of light passes through the focus of convex lens then it becomes parallel to the principal axis after refracting through the convex lens

In the given diagram, the portion BO of straight stick is immersed in water, but it appears to be bent at point B. The ray OC and OD coming from lower end of stick gets refracted away from normal in direction CX and CY. These rays when produced backwards appear to meet at ‘I’ so the virtual image BI is visible to us.

Thus, when a stick is half immersed in the water, then it appears bent at the surface due to refraction of light when light passes from water to the air.

(a) Convex mirrors are used as rear view mirrors in vehicles. It is due to the reason that convex mirrors produce an erect image of object and image formed in convex mirror is much smaller in size than the object due to which driver is able to see large area behind him and drive carefully.

(b) Concave mirrors are used as reflectors in search-light because when a light is placed at focus of concave reflector then it produces powerful beam of parallel rays, by which we can see things even in very dark.

There are two laws of refraction.

First law of Refraction of light:

The incident ray, refracted ray and the normal at the point of incidence, all lie in same plane.

In given Figure ray AO, OB and ON all lie in same plane i.e. plane of paper.

Second law of Refraction of light: The ratio of sine of angle of incidence (sin i) to the sine of angle of refraction (Sin r) is constant for a given pair of media.

Where ‘i’ is the angle of incidence and ‘r’ is the angle of refraction.

This constant is called refractive index. It has no units.

Given,

Object distance u = -15 cm

Focal length of convex lens f= 20 cm

Image distance v=?

We know that lens formula is:

Substituting the values in this formula

v = -60 cm

i.e.

Height of image = 4×height of object

● Image is formed on the same side of the object above the principal axis.

● It is virtual and erect.

● It is enlarged or magnified i.e. 4 times to the size of object.

when a ray of light passes through the focus of a concave mirror then it becomes parallel to the principal axis. Here incident ray AE is passing through the focus F of concave mirror and strikes the mirror at E, then it gets reflected through path EG parallel to the axis.

Also,

Angle of incidence = angle of reflection

∠i = ∠r

The relation between object distance, image distance and focal length of a spherical mirror is given by a formula called MIRROR FORMULA. It can be written as:

Where,

v is the distance of image from the mirror

u is the distance of object from the mirror

f is the focal length of the mirror

Mirror formula can be applied to all type of mirrors.

When object AB is placed between pole P and focus F then the image is formed behind the mirror which is virtual and erect (same side up as the object).

The image is larger in size than the object i.e. magnified.

Light ray AD which is parallel to axis gets reflected at D and passes through focus F and ray AE passes through centre of curvature after striking the concave mirror. Both reflected rays EC and DF are diverging. When these reflected rays are produced backwards, they intersect at A’ behind the concave mirror. Then we draw A’B’ perpendicular to axis thus A’B’ is the virtual image of the object.

If the image of the head is bigger than it must be a concave mirror. Concave mirror produces virtual, erect and enlarged image behind the mirror when object is placed between its pole and focus.

If the middle portion of her body is of the same size then it must be a plane mirror which produces same size virtual and erect image as object.

Since the image of leg is smaller than the leg itself, it must be convex mirror because convex mirror produces diminished image of the object.

So, the order of combinations for the magic mirror from the top.

1. Concave mirror

2. Plane mirror

3. Convex mirror

Object distance = 40 cm

Image: real, inverted and same size as object.

Real and inverted image is formed by the convex lens only. For image to be of same size as object, object must be placed at 2f distance from pole.

Since, object distance = 40 cm

2f = 40 cm

f = 20 cm

f=0.20m

Power of lens = 5 D

Given,

Image is real i.e. it is formed on same side of object, it is inverted and magnified so it must be concave mirror.

Image distance v= -30 cm

Magnification = -2

Object distance = -15 cm

Object is placed to the left of the mirror so object distance U= -15 cm.

We know that mirror formula is:

f = -10 cm

If the object is moved 10 cm towards the mirror i.e. Object distance = -15-(-10)

= -5cm

Which means that object is placed between pole and focus of the concave mirror so Image will be

● Virtual and erect

● Enlarged

Given

n_xy = 2/3

Where n_xy= refraction index of a medium ‘X’ with respect to medium ‘Y’

We know,

Substituting given value in the equation:

. . . . . . . . equation 1

It is also given that, n_yz = 4/3

Where n_yz = the refractive index of medium ‘Y’ with respect to medium ‘Z’

Substituting the given value in formula:

_{. . . . . . . . . . .} equation 2

Refractive index of medium ‘Z’ with respect to medium ‘X’ will be given by:

Substituting values from equation 1 and 2

Refractive index of medium ‘Z’ with respect to medium ‘X’ is 9/8

The property of concave mirror to magnify the image of object is utilized for using them as shaving mirrors.

Concave mirrors are used as shaving mirrors to see enlarged image of the face and make the shave. When the face is held within the focus of a concave mirror then the enlarged image can be seen in the concave mirror.

Ray diagram ofLight passing through rectangular glass prism is given below

Light emerges from the glass slab in direction parallel with that in which it enters the glass slab. Perpendicular distance between incident ray and emergent ray coming out of glass salb is called lateral displacement.

Ray diagram of Light passing through triangular glass prism:

The incident ray PQ refracts towards the normal at surface AB of prism and then refracts again away from normal through surface AC. The emergent ray RS is not parallel to the incident ray as in case of rectangular glass slab.

Given,

Focal length of concave lens = -50 cm

= m

= -0.50 m

To find: Power of lens?

We know,

Power of lens = - 2 Dioptres.

Since the focal length of a concave lens is always negative so power of the concave lens is -2 diopters.

Given,

Height of object h₁= 10 mm

Height of image h₂= 5 mm

Image distance v= -30 cm

To find focal length f =??

Substituting values given in this formula:

Magnification = 0.5

So, the given information signifies that image formed by mirror is of small size than that of object. Also,

Substituting values in this,

Object distance = 60 cm

Since object is placed on left of mirror so

u = -60 cm

We know that mirror formula is:

f = -20 cm

Focal length of the concave mirror is -20 cm.

Given,

Magnification= -1

Image distance v= 35 cm i.e. image formed behind the mirror so it is real.

Object distance u =??

Focal length =??

Height of image = - height of object

This means that the size of image is of same size as object but the image is inverted.

So it is a convex lens.

Also,

Object distance = -35 cm

Object is placed 35 cm left to the mirror i.e.

u = 35 cm

We know that lens formula is:

Substituting the values given:

f = 17.5 cm

If object distance is displaced 20 cm towards the optical centre of the convex lens then

Object distance =35-20 cm

= 15 cm

i.e. object is placed between pole and focus of convex lens

Image formed in this case will be:

● On same side of object

● Enlarged

● Virtual and erect

Given,

Focal length of convex lens = 18 cm

Object distance u = -27 cm

Image distance =??

We know that lens formula is:

Substituting the values given:

v = 54 cm

Since v is positive it means image is formed behind the lens. So, the image is real.

magnification = -2

Height of image = -2 ×height of object

Thus image formed is twice the size of object but –ve sign indicates that it is inverted.

a. object is placed at 30 cm from the

mirror.

Here focal length PF = 15 cm

Object distance PC = 30 cm

PC = twice the focal length = 2×15

When the object is placed at 30 cm from pole i.e. at the centre of curvature, an incident ray parallel to principal axis reflects at mirror and passes through the focus, Second ray of light passing through focus reflects at mirror and passes parallel to the principal axis. These both reflected ray DA’ and EA’ meet at a point A’. We draw A’B’ perpendicular to the principal axis which is the real and inverted image of the object.

Image is of same size as object.

A’B’ = -AB

Image distance v = 30 cm

b. object is placed at 10 cm form the mirror

We are given that focal length f is 15 cm, so if object distance is 10 cm then object must be between the pole and the focus.

In the given figure,

Object distance is BP = 10 cm

PF = 15 cm

Light ray AD from the object which is parallel to axis gets reflected at D and passes through focus F and ray AE passes through centre of curvature after striking the concave mirror. Both reflected rays EC and DF are diverging. When these reflected rays are produced backwards, they intersect at A’ behind the concave mirror. Then we draw A’B’ perpendicular to axis thus A’B’ is the virtual image of the object.

We know that mirror formula is:

B’P = v = -30 cm

Here image distance is 30 cm behind the mirror. The image formed will be virtual and erect.

Magnification = 3

A’B’ = 3 × AB

So, image is enlarged 3 times to the size of the object.

Given,

Focal length of concave mirror = 20 cm

If H is the height then,

H _image = 2× H _object

Magnification = 2

v = -2u . . . . . . equation 1

We know that mirror formula is:

Substituting the given values

2u = 20

U = 10 cm

So the position of the object is 10 cm to the left of the mirror

Given,

Height of object h₁= 2cm

Object distance u= -20 cm

Focal length of concave mirror f= -10 cm

To find:

Distance of screen = Image distance v=???

Height of image h₂=???

We know that mirror formula is:

Substituting the given values in the formula:

v = -20 cm

Screen should be placed at distance 20 cm from mirror in order to obtain sharp image

In other words, screen should be placed at distance 20+20 = 40 cm from the object in order to obtain sharp image.

Image will be formed in front of the mirror, i.e. real.

Also,

Height of image = - (Height of object)

This means that image is of same size as the object but it is inverted.

Given,

Object distance u₁ = -60 cm

Magnification m₁ = 1/2

If magnification m₂ = 1/3

Then object distance u₂ =???

v₁ = 30 cm

We know that mirror formula is:

Substituting the given values in the formula:

So , f = 60 cm

The image must have same focal length so

Substituting the given values here:

60 v₂ + 60 u₂ =

60 v₂ – u₂v₂ = -60 u₂

v₂ (60-u₂) = -60 u₂

We know that,

Substituting the above given values here:

60 – u₂ = 180

– u₂ = 180 – 60

– u₂ = 120

u₂ = -120 cm

The object should be placed at 120 cm from the mirror to get a magnification of 1/3

Given,

Focal length of concave lens f = -20 cm

Height of object h₁= 5 cm

Image distance v = -15 cm

Object distance u =???

Height of image h₂ =???

We know that lens formula is:

Substituting the given values in the formula:

u = -60 cm

Substituting the value object height here

Height or size of the image = 1.25 cm

Given,

Focal length of convex lens f= 18 cm

Image distance v= 36 cm

We know that lens formula is:

Substituting the given values in formula

u = -36

Substituting the values in the formula:

Given,

Object distance u= 8 cm

Focal length of A = 10 cm

=

= 0.10 m

Focal length of B = -10 cm

=

= -0.10 m

We know,

Power of lens A = 10 Dioptres.

Since the power of lens A is positive so it is a convex lens.

Similarly,

Power of lens B= -10 Dioptres.

Since the power of lens b is negative so it is a concave lens.

We know that lens formula is:

Substituting the values of lens A in formula

Substituting the values of lens B in above lens formula:

Thus, lens A which is convex will form a virtual and magnified image of an object placed 8 cm from the lens i.e. between focus and pole of lens.

(i) Ray diagram when object is placed in front of a convex lens at F

The image in this case is formed at infinity, it is real, inverted and highly enlarged.

(ii) Ray diagram when object is placed at 3F of a convex lens

Position of image in this case is between F and 2F on other side of lens, the image formed is real, inverted and diminished.

(iii) Ray diagram when object is placed between the lens and infinity of a concave lens

The position of Image in this case is between optical centre and the focus on the side of object. It is virtual, erect and diminished.