Light : Reflection And Refraction

Option (A): it is a fact that Wavelength of visible light is between 4 × 10 ^{- 7} and 8 × 10 ^{- 7}. Therefore, Option A is correct.

Radius of curvature (R) is equal to twice the focal length that means R = 2f. Therefore, Option C is correct.

Surface of Book is rough, therefore the reflection from book is irregular because the reflected light beam will not go in the particular direction.

Since book has rough surface therefore Option B is correct.

When ray of light passes through the centre of curvature of the concave mirror it strikes along the normal that means it is incident on mirror at 90⁰.Hence the incident ray coincides with the normal and retraces its path i.e. passes through centre of curvature.

As other options don’t go by above fact, so they are incorrect.

when an object is placed between the focus and pole of concave mirror the image is formed behind the concave mirror and is virtual, erect and enlarged. As shown in figure below.

Option (A): When object is kept at centre of curvature than image is formed at centre of curvature and is real, inverted. Therefore, Option A is incorrect.

Option (B): when an object is kept beyond centre of curvature the image formed is real, inverted and diminished. Therefore, Option B is incorrect.

Option (C): when an object is placed between focus and pole than image is virtual and erect as explained in explanation. Therefore, Option C is correct.

Option (D): when an object is placed at focus than image is formed at infinity and is real. Therefore, Option D is incorrect.

Magnification =

And plane mirror forms the image of same size as that of object. (Height of Image = Height of Object).

Therefore, Magnification is 1

A plane mirror can be considered as a spherical mirror of infinite radius of curvature. And Focal length is half of radius of curvature therefore it is infinity.

In plane mirror Object distance from the mirror is equal to the image distance from the mirror.

Since Object distance from mirror = 2 m

⇒ Image distance from mirror = 2 m.

∴ Distance between image and object = 2 m + 2 m = 4 m.

when an object is kept at centre of curvature of convex lens same size, real and inverted image of an object is formed at centre of curvature as shown in figure below.

Option (A): when an object is kept at focus than image is formed at infinity and is magnified. Therefore, Option A is incorrect.

Option (B): when an object is kept between focus and centre of curvature than image is formed on other side of lens and is magnified (enlarged). Therefore, Option B is incorrect.

Option (C): when an object is kept at centre of curvature than real, inverted image is formed at centre of curvature as explained in explanation. Therefore, option C is correct.

Option (D): when an object is kept between optical centre and focus the image formed is virtual and erect. Therefore, Option D is incorrect.

optical density is defined as the degree to which the refractive medium refracts the transmitted ray of light. Therefore, Medium having highest refractive index will have highest optical density.

For refractive indices,

Water (1.33) < pearl (1.46) < glass (1.52) < diamond (2.42)

Absolute refractive index of medium is defined as , and speed of light in vacuum is highest than its speed in any medium. Therefore, n>1.

Power of lens = .Therefore if focal length is less power is more and if focal length is more power is less.

Option (A): 50 cm is the highest focal length in given options; hence it will have lowest power. Therefore, Option A is incorrect.

Option (B): 25 cm is not the lowest focal length; hence it will not have highest power. Therefore, Option B is incorrect.

Option (C): 20 cm is not the lowest focal length; hence it will not have highest power. Therefore, Option C is incorrect.

Option (D): 10 cm is the lowest focal length in given options. Hence it will have highest power. Therefore, Option D is correct.

Power of lens = .

⇒ Focal length =

1m = 100 cm.

⇒ 0.2 m = 0.2 × 100 = 20 cm.

∴ Focal length = 20 cm. (Positive for convex lens.)

Option (A): since focal length is positive for convex lens. Therefore, Option A is incorrect.

Option (B): since focal length is positive for convex lens. Therefore, Option B is incorrect.

Option (C): focal length is 20 cm not 10 cm therefore option (C) is incorrect.

Option (D): as explained in explanation focal length = + 20 cm therefore option D is correct.

The relative refractive index between a pair of media is the ratio of their absolute refractive indices.

Let n₁ = absolute refractive index of water; n₂ = absolute refractive index of benzene; n₃ = absolute refractive index of sapphire.

Thus,

It can be seen that n₃₁ is maximum and corresponds to refractive index of sapphire relative to water.

Option (A): Plane mirror never forms real and inverted image, therefore option (A) is incorrect.

Option (B): the image formed by plane mirror is virtual and erect it is not real; therefore option (B) is incorrect.

Option (C): the plane mirror always form the virtual, erect image of same size as that of an object, therefore (C) is the correct answer.

Option (D): image formed is erect not inverted; therefore (D) is not the correct answer.

We know that

∴

×

⇒

⇒

optical density is defined as the degree to which the refractive medium refracts the transmitted ray of light. Therefore, Medium having highest refractive index will have highest optical density.

Option (A): Concave mirror and convex lens

Forms both virtual as well as real images, hence A is not the correct answer.

Option (B): for all positions of object convex mirror and concave lens always forms virtual and erect images behind the mirror/lens, therefore B is the correct answer.

Option (C): convex lens not forms only virtual images it also forms real images, hence C is not the correct answer.

Option (D): Concave mirror forms both real and virtual images, hence D is not the correct answer.

since the light ray is incident normally to the surface, therefore it will pass without any refraction. Hence angle of refraction = 0⁰.

For the compound microscope focal length of an objective lens is smaller than eye piece lens.f₀ = objective lens focal length, f_e = eye piece focal length.

⇒ f₀ < f_e

Regular reflection: when a parallel beam of light is incident on a plane or smooth surface, a beam remains parallel after reflection in a specific direction. Such reflection is called regular reflection. The figure below illustrates regular reflection.

Example: Reflection of light by plane mirror.

Irregular reflection: when a parallel beam of light is incident on rough or irregular surface like Book, stone etc., the beam does not remains parallel but spreads over wide region after reflection. Such reflection is called irregular reflection. The figure below illustrates irregular reflection.

Example: we can see book, table, and chair etc. due to irregular Reflection of light these objects.

the laws of reflection of light is as follows.

The angle of incidence is equal to the angle of reflection that means .The figure below illustrates it.

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

The figure below illustrates the law more clearly.

Centre of curvature: the centre of a spherical shell from which the mirror is made is called as the Centre of curvature of mirror. It is denoted by ‘C’.

Radius of curvature: the radius of a spherical shell from which the mirror is made is called as the radius of curvature of mirror. It is denoted by ‘R’.

The figure below shows the centre and radius of curvature of concave lens.

Ray Diagram:

Position of Image: Between Centre of curvature (C) and focus (F).

Nature of image: real and inverted.

Size of image: diminished.

Ray Diagram:

Position of Image: behind the mirror

Nature of image: virtual and erect

Size of image: Magnified (enlarged).

Ray Diagram:

Position of Image: between pole (P) and principal focus (F) behind the mirror.

Nature of image: virtual and erect.

Size of image: diminished.

In case of plane mirror the magnification is + 1;

Positive value of magnification indicates that image formed by plane mirror is erect.

Also, Magnification =

⇒ Image height = Object height.

Therefore, Image formed by plane mirror is of the same size as of the object.

Magnification =

⇒ v = - u;

This shows that image formed by plane mirror is at same distance as the object but behind the mirror means image is virtual.

Thus, Image formed by plane mirror is virtual, erect and same as size of the object.

The refraction of light occurs according to the following laws which are as follows.

The incident ray, refracted ray, and the normal to the surface separating two media at the point of incidence all lie in the same plane.

The ratio of sine of angle of incidence and sine of angle of refraction remains constant ( = constant) subject to certain conditions. This law is called Snell’s law of refraction.

The refractive index of medium with respect to the vacuum is called absolute refractive index.

Snell’s law in terms of refractive indices of two media: Let,

N₁ = absolute refractive index of medium 1.

N₂ = absolute refractive index of medium 2.

V₁ = velocity of light in medium 1.

V₂ = velocity of light in medium 2.

C = velocity of light in vacuum.

Then

N₁ = and N₂ =

∴

⇒ N₂₁ = ………….. (1)

Now According to Snell’s law

N₂₁ = …………….. (2)

From (1) and (2)

This is the general form of Snell’s Law.

Ray Diagram:

Position of Image: on other side of lens at centre of curvature.

Nature of image: real and inverted

Size of image: Same as size of object.

Ray Diagram:

Position of Image: beyond the centre of curvature (2F) on the same side of lens where object is kept

Nature of image: virtual and erect

Size of image: Magnified (enlarged).

Ray Diagram:

Position of Image: between focus (F) and optical centre (C) on the same side of lens where object is kept.

Nature of image: virtual and erect

Size of image: small (diminished).

Magnification of an image is defined as the ratio of the height of image to the height of object.

Magnification (m) =

In the figure above by geometry we can easily see that ΔABO and ΔA’B’O’ are similar.

∴ …………… (1)

From definition of magnification.

Magnification (m) =

⇒ ……… (2)

From (1) and (2)

Magnification = which is the formula for magnification of lens.

MM’ = plane mirror.

AB = object of size ‘h’ on the left - hand side of the mirror at a distance ‘u’.

AP is the incident ray of light from point A of the object which falls on the mirror at point P. This incident ray is reflected back in the same path PA as shown in the figure above.

Another incident ray OC falls on the mirror at a point O and is reflected along the path OC.

Now, since reflected rays PA and OC are diverging and therefore cannot meet each other in front of the mirror, hence we extend these rays behind the mirror by dotted lines as shown in the figure below.

On extending these rays behind the mirror, we see that these rays meet at point A’ at a distance V’. Therefore A’ is the virtual image of point A of the object AB.

Similarly, a virtual image of point B will be formed behind the mirror as B’.

Now, to get a complete image of the object AB, we join the point A and B to point A’ and B’ by a dotted line.

We find that the image A’B’ being formed is virtual, erect and of same shape and size as the object AB; thereby giving us the characteristics of images formed by the plane mirror.

Pole (P) of the mirror is taken as the origin and the Principal axis is taken as the horizontal X - axis. The sign conventions are as follows:

The object is always placed on the left side of the mirror.

All the distances measured to the right side of the origin are taken as positive, while the distances measured to the left side of the origin are taken as negative.

All the distances parallel to the principal axis of the mirror are measured from the pole of the mirror.

Distances measured vertically upwards from the principal axis of the mirror are taken as positive.

Distances measured vertically downwards from the principal axis of the mirror are taken as negative.

The focal length of Concave mirror is negative while that of convex mirror is positive.

The figure below illustrates the conventions used.

Consider the rectangular slab as shown in the figure below.

AE = incident ray;

i = Angle if incidence;

r = Angle of refraction;

On entering the glass slab the light ray bends towards the normal and travels along EF.

The refracted ray EF is incident on the face SR at angle of incidence r’.

The emergent ray FD bends away from normal at angle of refraction e.

The emergent ray FD is parallel to the incident ray AE but it has been laterally displaced with respect to incident ray.

Consider the image formation by convex lens as shown in the figure below.

By Cartesian sign convention;

Object distance (OB) = - u;

Image distance (OB’) = + v;

Focal length (OF₂ = OF₂) = f

From the geometry we can see that in the figure above right triangle ΔABO and ΔA’B’O’ are similar.

∴ …… (1)

Also, in figure above by geometry right triangle ΔODF₂ and ΔBAF₂ are similar.

∴

∴ (∵ OD = AB, opposite sides of rectangle ABOD are equal).

∴

∴ ………. (2)

From (1) and (2)

⇒

⇒ .

Dividing each term by uvf, we get.

⇒

⇒

This equation is called the ‘Lens Formula’

The image formed by spherical mirrors can be located by constructing ray diagrams. The point of intersection of two rays will give the position of the image.

Following rays can be taken into consideration while locating the image.

1. A ray parallel to the principal axis after reflection passes through the focus or appears to diverge from the focus. As shown in the figure below.

2. A ray passing through the principal focus of the concave mirror or ray which is directed towards the principal focus of convex mirror will emerge parallel to the principal axis as shown in the figure below.

3. A ray passing through the centre of curvature of the concave mirror or directed towards the centre of curvature of convex mirror after reflection is, reflected along the same path, as shown in the figure below.

4. A light ray incident obliquely to the principal axis towards the pole of a concave mirror or convex mirror is reflected obliquely following the laws of reflection. The figure below shows it.

Power of lens is defined as the efficiency with which a lens can converge or diverge the light ray.

The reciprocal of focal length of the lens is called the power of the lens (P).

P (in dioptre) =

The S.I unit of power of lens is dioptre denoted by D.

Power of convex lens is taken as positive and of concave lens is negative. The instrument used for measuring the power of lens is called dioptremeter.

According to the Cartesian system in the above figure.

Object distance (PB) = - u

Image distance (PB’) = - v

Focal length (PF) = - f

Radius of curvature (PC) = - R

From the geometry we can see that in the figure above right triangle ΔABP and ΔA’B’P’ are similar.

∴

∴ ……. (1)

Also ΔABC and ΔA’B’C’ are similar.

∴ ……. (2)

Also

From above figure

CB’ = PC – PB’ = - R - (- v) = - R + v.

And

CB = PB – PC = - u - (- R) = - - - u + R.

From (2)

∴

⇒ R (u + v) = 2uv

Dividing both sides by Ruv we get.

We know that f =

Putting this in the above equation we get

This equation is the ‘Mirror Formula’

Construction: A compound microscope consists of two convex lenses. The lens towards object is called objective lens and the lens near eye is called the eye piece. The focal length of objective lens is less than the focal length of eye piece. This combination forms the magnified image of the object.

The figure below shows the compound microscope.

Working: The object AB is placed in front of objective lens in such a way that its real, inverted and magnified image A’B’ is obtained beyond the centre of curvature of objective lens.

The image A’B’ becomes an object for the eye piece. The position of image is adjusted such that it lies within the focal length of eye piece.

The eye piece forms the virtual, erect and magnified image A”B” of the object.

An astronomical telescope is an optical instrument which is used to see the magnified image of distant heavenly bodies like stars, planets, satellites and galaxies etc.

The final image formed by an astronomical telescope is always virtual, inverted and magnified.

Principle: An astronomical telescope works on the principle that when an object to be magnified is placed at a large distance from the objective lens of telescope, a virtual, inverted and magnified image of the object is formed at the least distance of distinct vision from the eye held close to the eye piece. Figure below shows the astronomical telescope.

Working: Parallel rays of light coming from the distant object forms real, inverted and diminished image of an object.

This image acts as an object for the eye piece. The eye piece than forms virtual, erect and magnified image of the distant object.

According to the question;

Object distance (u) = - 10 cm;

Focal length (f) = 15 cm Image distance = v cm;

Height of object = 5 cm

By Mirror Formula;

⇒ .

⇒

⇒

⇒ V = 6 cm.

Since image distance v is positive

⇒ Image is virtual and erect.

Magnification (m) =

⇒

⇒

∵ v < u ⇒ Magnification < 1 hence image is diminished.

A virtual, erect and diminished image of height 3 cm is formed 6 cm behind the mirror.

According to the question;

Object distance (u) = - 15 cm;

Focal length (f) = - 10 cm Image distance = v cm;

Height of object = 6 cm;

By Mirror Formula;

⇒ .

⇒

⇒

⇒ V = - 30 cm.

Since image distance v is negative

∴ Image is real and inverted.

Magnification (m) =

⇒

⇒

∴ Height of image = 12 cm.

∵ v > u ⇒ Magnification > 1 hence image is magnified.

A real, inverted and magnified image of height 12 cm is formed 30 cm in front of the mirror beyond centre of curvature.

Refractive index (R.I) of glycerine with respect to glass =

⇒ Refractive index (R.I) of glycerine with respect to glass =

Hence Refractive index (R.I) of glycerine with respect to glass is 0.98

Refractive index (R.I) of glass with respect to water =

⇒ 1.12 =

⇒ Absolute R.I of water =

Hence Absolute Refractive index of water is 1.34

Refractive index (R.I) of water with respect to glass =

Given;

Angle of incidence (i) = 26^o48’

Refractive index of water with respect to glass = 0.9

⇒ 0.9 =

⇒ =

⇒

⇒ r =

⇒ r = 30⁰.

Hence angle of refraction is 30⁰.

According to the question;

Focal length (f) = + 10 cm

Object distance (u) = - 15 cm.

Image distance = v.

By lens formula;

⇒

⇒

⇒

⇒ V = 30 cm.

Hence image is formed at distance of 30 cm from the lens.

According to the question;

Focal length (f) = - 30 cm

Object distance (u) = - 20 cm.

Image distance = v.

By lens formula;

⇒

⇒

⇒

⇒ V = - 12 cm.

Hence image is formed at distance of 12 cm from the lens.

Given;

Power of convex lens = + 4 D.

We know that;

Power (P) =

∴ Focal length (f) =

Hence f = 0.25 m.

Also

Given that;

Size of the object = size of the image.

For convex lens this is possible only when the object is placed at the centre of curvature.

∴ Object distance = 2f = 2 × 0.25 = 0.5 m.

0.5m = 0.5 × 100 = 50 cm.

Hence object is placed at 50 cm at centre of curvature.