Light: Reflection And Refraction

When ray of light passes through the center of curvature of the concave mirror it strikes along the normal that means it is incident on mirror at 90°.

Thus by the law of reflection which states that the angle of incidence is always equal to the angle of reflection i.e. 90°.

Hence the incident ray coincides with the normal and retraces its path after reflection from the spherical mirror.

The figure below illustrates the above situation.

Image formed is real and erect because real image is formed on the screen, And it is not inverted because the object inside the projector is inverted already, therefore its image will come as erect image, hence we are able to produce the image on screen as well as keep it erect.

A mirror forms an image such that its left side is object’s right side and its right side is object’s left side. This phenomenon of change in object and image sides is known as lateral inversion.

For example, the word AMBULANCE is painted left-right inverted on the ambulance so that when the driver of a vehicle in front looks into his rear-view mirror, he can make out the word AMBULANCE quickly and give way. As shown in the figure below.

Convex mirror always forms virtual, erect and diminished image of an object but when the object is kept at infinity the image is formed at focus and is extremely diminished. As shown in the figure below.

Convex mirror has a larger field of view because it is diverging in nature. The light rays incident on it gets diverged following the laws of reflection, the reflected rays form the wider path as shown in the figure below.

Due to large field of view convex mirrors are used in headlights of vehicles.

Ray of light strikes the mirror at an angle of 30⁰.

⇒ It will make an angle of 60⁰ (90⁰ – 30⁰) with the normal. As shown in the figure below.

∴ Angle of incidence = 60⁰.

Now, By laws of reflection;

Angle of Incidence = Angle of reflection;

⇒ Angle of reflection = 60^o.

Hence Angle of reflection is 60^o.

The mirror is likely to be the concave mirror because the focal length of the concave mirror is negative because the focus is on the left-hand side of the mirror which is taken to be negative according to sign convention.

Thus the required spherical mirror is a concave mirror of focal length 10 cm.

Positive magnification means that both the object and the image formed by a spherical mirror (concave or convex) are erect (upright). An erect image is always virtual. Thus, positive magnification indicates that the image formed by a spherical mirror is virtual, erect.

Convex mirrors are used in vehicles for rear view due to their wide field of view. And the convex mirror forms a virtual, erect and diminished image of the object. Therefore, magnification (m) produced by a rear view mirror fitted in vehicles is less than one i.e., m < +1 because

Magnification (m) =

And image size is less than the object size.

The rays from sun coming from infinity are parallel to principal axis after reflection they converge at a point, known as focus. Therefore, focal length (F) of concave mirror is 15 cm.

And we know that same size, real and inverted image is formed by concave mirror when object is placed at centre of curvature = 2×Focal length (F). So to form same size image object will be placed at 15x2 = 30cm.

According to the question;

Focal length (f) = 12cm;

Object distance (u) = -8cm;

By mirror formula;

⇒

⇒ v = .

Since v = 4.8cm which is positive hence image is behind the mirror.

The image is formed at the distance of 4.8 cm behind the mirror and is virtual and erect.

Given that;

Object distance (u) = -24cm;

Image distance (v) = -12cm (image is in front of mirror hence v is negative).

Now;

Magnification =

⇒ Magnification =

∴ Magnification is -0.5. Which tells that image is real and inverted and half of the size of the object.

The depth appears to be less because of refraction of light. Light rays bend as they move from water to the air. When the light ray from water reaches our eye, the eye traces them back as straight lines (shown as dashed lines) which intersect at a higher position than where the actual rays originated. This causes the depth of a bucket to appear higher and we cannot see the actual depth of the bucket.

The figure below illustrates the same.

Ray of light strikes the mirror at an angle of 60^o.

⇒ It will make an angle of 300 (90⁰ – 60⁰) with the normal.

∴ Angle of incidence = 30⁰.

Now, By laws of refraction;

Angle of incidence is 60^o

Refractive index of glass slab is 1.5

Putting the value in the above equation, we get

On solving the above equation, we get

Angle of refraction, r= 35.26^o

Thus the angel of refraction is 35.26^o from the normal.

Refractive index compares the speed of light in a particular medium with the speed of light in vacuum

Relation between refractive index and speed of light:
refractive index (μ) =

⇒

Let the refractive index of medium 1 be μ1 and the speed of light in medium 1 is v1.

⇒ ………………….. (1)

Similarly;

Let the refractive index of medium 2 be μ2 and the speed of light in medium2 is v2.

⇒ ………………….. (2)

Dividing (1) by (2)

⇒

Hence refractive index of a medium is inversely proportional to the speed of light in that medium.

Refractive index (μ) = ……(1)

Hence refractive index of a medium is inversely proportional to the speed of light in that medium.

Also μ = …………….. (2)

⇒ Refractive index is inversely proportional to the angle of refraction.

From (1) and (2)

Velocity of light is directly proportional to the angle of refraction.

Thus velocity of light is maximum in medium C because angle of refraction is 40⁰ is maximum.

Given;

Angle of refraction(r) = 300.

Angle of incidence = i.

Refractive index (μ) = √3.

Refractive index (μ) =

⇒ √3 =

⇒ √3 = 2×sin i

⇒

⇒

Hence angle of incidence is 60°.

Given;

Refractive index of alcohol (μ) = 1.36.

Speed of light in vacuum = 3×10⁸m/s.

Refractive index (μ) =

⇒ 1.36 =

⇒ .

∴ speed of light in alcohol = 2.21 × 10⁸m/s

speed of light in alcohol with respect to air is 2.21 x 10⁸ ms^-1

Refractive index of air, μair = 1;

Given;

Refractive index of water w.r.t air, aμw ;

Refractive index of glass w.r.t air, aμg ;

Refractive index of water w.r.t glass = gμw = =

⇒ gμw =

∴ Refractive index of water w.r.t glass is .

A concave lens always forms the virtual image of an object because it is diverging lens. The image formed is erect as shown in the figure below.

A virtual, erect and enlarged image is formed at the same side of lens, when an object is placed between principal focus, F1 and optical centre, O of a convex lens.

Figure below shows the image formation.

I would prefer convex lens because it gives the magnified image of an object when an object is placed between the radius of curvature and focal length. Also the magnification is more for the convex lenses having shorter focal length.

Hence for reading, a convex lens of focal length 5 cm should be used.

When the light ray passes from a rarer medium to a denser medium it bends away from the normal. Thus light ray get diverge when it emerges out from the air bubble.

Thus the air bubble inside the water acts as Concave lens.

Yes the above statement is correct.

A magnified virtual image can be obtained when the object is placed between optical centre and principal focus of the lens means the object is placed at distance less than 20 cm from the centre of the lens. Figure below illustrates it.

Real image of various sizes can be obtained when the object is placed beyond the principal focus.

Yes the above statement is correct.

A magnified virtual image can be obtained when the object is placed between optical centre and principal focus of the lens means within 20cm. Figure below illustrates it.

Real image of various sizes can be obtained when the object is placed beyond the principal focus.

Since the focal length is negative, therefore the lens is concave lens because the focal length of concave lens is negative.

P (in diopter) =

F = -50cm = -0.5m (1m = 100cm)

⇒ P =

Hence power of lens is -2 Dioptre.

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

Also P (in diopter) =

Hence power is inversely proportional to the focal length

Therefore the lens with focal length 20cm will have more power than the lens with focal length 30cm. Therefore the lens with focal length 20cm will give more convergent light.

Given Focal length (f) = 4D

P (in diopter) =

⇒ 4 =

⇒

Now 1m = 100cm

⇒ 0.25m = 25cm

Hence the focal length of lens is 25cm.

Power P1 = -3.5D; P2 = +1D

Total power = P1 + P2 = -3.5D + 1D

⇒ Total power (P) = -2.5D.

Now;

P (in diopter) =

⇒ -2.5 =

⇒

⇒

Now 1m = 100cm

⇒ -0.4m = -40cm

Hence the focal length of lens is -40cm.

If the two lenses are kept in contact than their power is added

⇒ Power (P) = P₁ + P₂.

Hence the power of combination is increased which means that the converging capacity is increased. And the focal length is decreased (because focal length is inversely proportional to the power of lens).

Therefore the ray after refraction will converge at point close to lens than earlier.

(i) If the lens is cut along the principal axis than there will be no change in the focal length of the lens because the radius of curvature will remain the same, hence the focal length will remain unchanged.

(ii) If the lens is cut perpendicular to the principal axis than the focal length will change it will be twice the focal length of the original lens because the radius of curvature will increase and R = 2F.

A plane mirror can be considered as a spherical mirror of infinite radius of curvature where the radius is infinitely large hence the aperture is straight. Therefore the radius of curvature of a plane mirror is infinite

The sunglasses are not used for magnification they are used to protect our eyes from the sun rays. The sun glass is the plain glass with thin polarizing film, therefore it does not have focus or the focus is at infinity.

∴ Power =

Power =

Hence the power of sun glasses is zero.

The refractive index of free space is 1, hence μ = 1.

⇒

⇒ Speed of light = c

Therefore speed of light in free space is 3×10⁸m/s.

When an object is kept at the centre of curvature of the concave mirror than the image formed is real, inverted and equal to the size of the object.

The figure below shows the ray diagram.

Yes it does change with the change in medium because different mediums have different optical densities. Some are optically denser(higher refractive index) like oil, diamond etc in which the light travels at a slower speed and bends towards the normal, whereas some other materials can be optically rarer(lower refractive index) like air, water in which light speeds up and bends away from the normal.

When an object is kept at infinity or larger distances than the image is formed at the focus. Therefore the image of the distant tree will be formed at the focus of the concave mirror. So the screen will be placed at the focus to obtain the image And the distance between the mirror and the screen will be the focal length.

Figure below shows the ray diagram for the same.

Yes the refractive index changes with the color of the light because refractive index changes with the change in the wavelength and different colors have different wavelength. Due to it refractive index of violet color is highest and of red color is lowest.

Given;

Magnification (m) = 0.16;

Size of object = 4cm;

Size of image =?

We know that;

⇒

⇒ Size of image = 0.16×4 = 0.64cm.

Hence size of image is 0.64cm.

When the ray of light passes through the focus of the convex lens it becomes parallel to the principal axis as shown in the figure below.

Given that;

Refractive index (μ) = 1.33 ;

Speed of light in air(c) = 3 x 108;

Speed of light in water =?

We know that

⇒

⇒

⇒ v= 2.25 × 10⁸m/s.

Hence speed of light in water is 2.25 × 10⁸m/s.

A ray of light parallel to the principal axis after reflection appears to come from the focus of the convex mirror as shown in the figure below.

A medium can be classified as rarer or denser by following ways.

Rarer medium has less refractive index ex-water (1.33) while denser medium has higher refractive index ex-diamond (2.42).

When light goes from rarer medium to denser medium it bends towards the normal but when light goes from denser to rarer medium it goes away from the normal.

The matchstick will burn because convex lens is the converging lens. The rays of sun coming from large distance ( infinity) will be converged at the focus of the convex lens, due to which too much energy will be produced and this energy will burn the matchstick kept at focus of the convex lens.

Figure below shows the ray diagram for the same.

Given;

Magnification (m) = 4;

Object distance (u) = -4cm;

Image distance = v;

Magnification (m) =

⇒

⇒ v = 4 × 4 = 16cm

Image distance = 16cm.

By mirror formula;

⇒

⇒

⇒

⇒

Hence the focal length is -5.33cm. Since the focal length is negative therefore the mirror is concave.

Since the ray strikes the glass slab at an angle of 90°. Therefore

Angle of incidence = 0⁰.

Thus the ray of light will pass straight without any deviation as shown in the figure below.

Laws of Refraction of light;

Refractive index=1.5

Thus angle of refraction is given by

On solving, we get

Thus the angle of refraction is 0^o

Thus the refracted ray is perpendicular to the surface of the glass slab as shown in the above diagram.

When two or more lenses each of power P₁, P₂, P₃, P₄… P_n . are placed in contact with each other as shown in the figure below.

Then the combined power of the combination is the sum of all individual powers.

⇒ P = P₁ + P₂ + P₃ + P₄ + …..+ P_n

It happens so due to the refraction in glass Slab. The incident ray bends towards the normal as the refracted ray when it enters the glass slab, and bends away from the normal as it comes out of the slab as the emergent ray. Because of the displacement between the incident ray of light and the emergent ray of light. The displacement between emergent and incident ray is called as lateral displacement.

The figure below illustrates the above situation.

When ray of light passes through the centre of curvature of the spherical 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. Therefore angle of incidence = 0 so, According to law of reflection;

Angle of reflection = 0, hence the angle of reflection to become zero degrees, thus ray of light retraces its path.

The figure below illustrates the above situation.

Bending of the light ray as it travels from one medium to another is due to the refraction of light. Refraction takes place because there is change in velocity as the light travels from one medium to another. Also the light always takes the shortest path between the two given points, due to this property the light gets bend.

When the ray of light travels from denser medium to the rarer medium than it bends away from the normal.

Therefore medium x is optically denser than medium y because the ray of light bends away from the normal when it travels from the medium x to medium y.

Figure below shows the situation.

As we move above the earth surface the density of air goes on decreasing because the amount of air present at high altitudes is less. Thus due to the decrease in the density of the air, the refractive index gradually increases from the atmosphere to the earth’s surface. Therefore the refractive index of the atmosphere is different at different altitudes.

Refractive index, ……(1)

Hence refractive index of a medium is inversely proportional to the speed of light in that medium.

Also μ = …………….. (2)

⇒ Refractive index is inversely proportional to the angle of refraction.

From (1) and (2)

Velocity of light is directly proportional to the angle of refraction.

Thus velocity of light is minimum in medium R because angle of refraction is 150 is minimum.

Refractive index is the measure of bending of ray of light as it passes from one medium to another. It is also defined as the ratio of the speed of light in vacuum to the speed of light in any medium.

Refractive index (μ) =

Given

Speed of light in a medium (v) =

⇒

⇒ μ =

⇒ μ =

Hence Refractive index of medium is 1.5.

Given;

Radius of Curvature(R) = 50cm;

R = 2f

⇒ 50 = 2f

⇒

Focal length = 25cm.

When an object is kept at the focus of the concave mirror then the image is formed at infinity.

This can be shown as below:

The Mirror formula is given by:

Where, F is the focal length of the mirror

v is the image distance from the pole of the mirror

u is the object distance from the pole of the mirror

On solving for v, we get

If F=u=25 cm, we get

Therefore the object should be placed at 25cm from the pole of the concave mirror to get an image at infinity.

The mirror is the concave mirror and the object is kept between Centre of curvature and focus.

Thus the real, inverted and enlarged image will be formed beyond the center of curvature. Below is the ray diagram for it.

According to the question;

Object distance = u;

Image distance (v) = 30cm;

Focal length = 20cm

By lens formula;

⇒ u = -60cm.

Therefore object is placed at 60 cm in front of lens.

Now;

Height of object h₁= 6cm;

Magnification = =

Putting values of v and u

Magnification =

⇒

⇒

Height of image is 3 cm.

Negative sign means image is real and inverted.

According to the question;

Object distance (u) = -27cm;

Focal length (f) = -18cm;

Image distance = v;

By mirror formula;

⇒

⇒ v = -54cm.

Thus, screen should be placed 54cm in front of the mirror to obtain the sharp focused image.

Height of object h₁= 7cm;

Magnification = =

Putting values of v and u

Magnification =

⇒

⇒ h2 = 7 ×-2 = -14.

Height of image is 14 cm.

Negative sign means image is real and inverted.

Thus real, inverted image of 14cm size is formed.

According to the question;

Object distance = -30 cm;

Image distance = v cm;

Focal length = 20 cm

By lens formula;

⇒ v = 60cm.

Therefore image is formed at 60 cm in right of lens.

Now;

Height of object h₁= 5cm;

Magnification = =

Putting values of v and u

Magnification =

⇒

⇒ h₂ =-2 ×5=10

Height of image is 10 cm.

Negative sign means image is real and inverted.

According to the question;

Object distance, u= -45cm;

Image distance, v= 90 cm;

Focal length, f=?

By lens formula;

⇒ f = 30cm.

Since the focal length is positive, therefore the lens is convex lens of focal length 30cm.

The incident ray and the emergent ray do not coincide they are parallel to each other. The emergent ray is shifted by some distance from the incident ray while passing throught the glass slab which is called the lateral displacement.

The figure below illustrates it.

Refractive index (μ) =

Given μ = 2.42, c = 3 × 10⁸.

Putting the value in the above formula, we get

⇒

⇒

⇒ v = 1.23 × 10⁸ m/s.

Speed of light in diamond is 1.23 × 10⁸ m/s

% of speed in diamond = = 41%

Thus there is 41% decrease in the speed of light in diamond than in air.

According to the question;

Object distance (u) = -60cm;

Image distance (v) = 120cm;

Focal length = f

By lens formula;

⇒ f = 40cm.

Therefore Focal length of lens is 40 cm.

Since the focal length is positive. Therefore the lens is convex lens of focal length 40cm.

Now;

Height of object h₁= 5cm;

Magnification = =

Putting values of v and u

Magnification =

⇒

⇒ h₂ = -2 × 5= -10 cm

Height of image is 10 cm.

Negative sign means image is real and inverted..

Power P₁ = 2D;

Since Power is positive. Therefore, the lens is convex lens.

⇒

⇒

1m = 100cm

⇒ 0.5m = 50cm.

Hence the focal length is 50cm

P₂ = -4D.

Since the power is negative. Therefore, the lens is concave lens.

⇒

⇒

1m = 100cm

⇒ 0.25m = 25cm.

Hence the focal length is -25cm

(a) Given

Object distance (u) = -100cm

Focal length (f) = 50cm

Image distance = v

By lens formula;

⇒ v=100cm

Hence the image is formed at 100 cm behind the lens.

Object distance (u) = -100cm

Focal length (f) = -25cm

Image distance = v

By lens formula;

⇒ v =

⇒ v= -20cm

Hence the image is formed at 20cm in front of the lens.

(b) For 1st case

Object distance (u) = -100cm

Image distance (v) = 100cm.

By Magnification Formula

Magnification (m) =

⇒ Magnification =

Therefore, Magnification for the 1st case is -1. The negative sign means the image is real and inverted.

For 2nd case

Object distance (u) = -100cm

Image distance (v) = -20cm.

By Magnification Formula

Magnification (m) =

⇒ Magnification =

Therefore, Magnification for 2nd case is 0.2. The positive sign means the image is erect and virtual.

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.

The object distance is always 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.

Given;

Magnification (m) = -3 (real image is inverted always).

Object distance (u) = -16cm.

Ray diagram can be shown as below

By Magnification Formula;

Magnification,

Where,

v is the image distance

u is the object distance

⇒

⇒ v = -3 × 16 = -48cm.

∴ Image distance (v) = -48cm.

By mirror formula;

Therefore the focal length is -12 cm. Since the focal length is negative, therefore the mirror is concave mirror.

According to the new Cartesian system the Pole (P) of the mirror is taken as the origin and the Principal axis is taken as the horizontal X- axis. The sign conventions for it 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.

Given;

Magnification (m) (virtual image is always erect).

Object distance (u) = -18cm.

By Magnification Formula;

Magnification,

Where,

v is the image distance

u is the object distance

⇒

⇒

⇒ v= 6cm.

∴ Image distance (v) = 6cm.

By mirror formula;

⇒

Therefore, the focal length is 9 cm. Since the focal length is positive, therefore the mirror is convex mirror.

Figure below shows the ray diagram for the same.

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 diopter denoted by D.

From the formula of power of lens.

1 D =.

Thus 1 Dioptre is the power of the convex lens having focal length of 1m.

Given

f₁ = 25cm (Focal length of convex lens is positive)

⇒ f₁ = 0.25m (1m = 100cm)

P (in dioptre) =

⇒ Power of convex lens,(F=25 cm) P₁ =

⇒ P₁ =
∴ P₁ = 4D

f₂ = -10cm (Focal length of concave lens is negative)

⇒ f₂ = -0.1m (1m = 100cm)

⇒ Power of the concave lens(F=10 cm), P₂ =

⇒ P2 =

∴ P2 = -10D.

Power of combination = P₁ + P₂.

P = 4D – 10D = -6D.

Lens power of the combination is -6D. Since it is negative therefore the new lens will behave like concave lens having a focal length of 16.6cm.

Sign convention for the refraction of light is as follows.

1. All the distances are measured from the optical centre of the lens.

2. All the distances which are measured in the same direction of the incident light ray (left hand side of lens) will be taken as positive.

3. All the distances which are measured against the direction of incident light (right hand side) are taken as negative.

4. The perpendicular distances to principal axis in upward direction will be positive and those in downward direction will be taken as negative.

Figure below illustrates the conventions more clearly.

According to the question.

Magnification (m) = -3 (real image is always inverted).

Object distance (u) = -16cm.

By Magnification Formula;

Magnification,

Where,

v is the image distance,

u is the object distance

⇒

⇒ v= -16 × -3

⇒ v= 48cm.

∴ Image distance (v) = 48cm.

By lens formula;

⇒

Therefore the focal length is 12 cm. Since the focal length is positive, therefore the lens is convex lens.

Figure below shows the ray diagram for the same.

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

1. 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.

2. 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.

Refractive index of a medium (μ) =

⇒ μ =

Refractive index of air, μ_air = 1;

Given;

Refractive index of medium x w.r.t medium y, μ_xy;

Refractive index of medium y w.r.t medium z, μ_yz;

Refractive index of medium z w.r.t medium x = μ_zx =

⇒ μzx =

⇒ μzx =

⇒ μzx =

⇒ μzx =

⇒ μzx =

∴ Refractive index of medium z w.r.t medium x is .

(i) Image formed by plane mirror is of the same size as of the object.

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

Image formed by plane mirror is at same distance as the object but behind the mirror.

Magnification of image is +1.

The image is virtual.

(ii) Varun made an arrangement for the light so that his mother could finish the work easily this shows that Varun is caring and responsible towards his mother. Also, he has used his knowledge of plane mirrors to get the light this shows that he is also quite clever and intelligent also he is the serious student who focuses on the practical aspect of things taught in class and applies them in a day to day use.

In other words, Varun had put the science to the real use.

(i) The convex mirrors are diverging in nature and they have the wide field of view due to which we can see the image of large area in the mirror. The light rays incident on it gets diverged following the laws of reflection, the reflected rays forms the wider path as shown in the figure below.

(ii) Mr. Raghav every time look to see whether students are drawing the diagram or not this shows that he is a responsible and caring teacher who does his job passionately. Arav puts two convex mirrors so that Mr. Raghav will not have to turn every time this shows that Arav is helpful, caring and respectful towards his teacher. Also he had used the property of convex mirrors thus he is quite intelligent and serious student who puts the practical use of concepts in day to day life use.

Given:
Magnification (m) = 3 (virtual image is erect).

Radius of the curvature, R =36 cm
Focal Length (f) = = -18cm (Negative for concave mirror).

Now by formula,

1. Magnification,

Where,

v is the image distance,

u is the object distance,

Putting the values in the above formula, we get

⇒

2. Mirror formula

Where
f is the focal length of the mirror
v is the image distance from the mirror
u is the object distance from the mirror
Thus, putting the values in the equation, we get

Thus, on solving the above equation, we get

u=-12 cm

Hence the distance of the object from the mirror is 12 cm.

According to the question

Magnification (m) = (image is of the size of object).

Focal length = f (positive for convex mirror).

Object distance = -u.

Now By formula.

⇒

⇒ f + u = nf

⇒ u = nf – f

⇒ u = (n-1)f.

Therefore the object distance from the mirror is (n-1)f. Hence the option C is correct.

Given: -

Area of the mirror = 1 cm²,

Power transmitted to the mirror = 10 Wcm^-2,

Calculations: -

The average force is given by ,

Since the surface is perfectly reflecting we can write

So

6.7×10^-8 N

When an object is placed at centre of curvature of concave mirror than the real, inverted image of size same as that of object is formed at the center of curvature.

Therefore the image distance = Centre of curvature = 2f. Hence the correct option is C

Refractive index of air, μ_air = 1;

Given;

Refractive index of flint glass w.r.t air, ^airμ_glass = 1.65;

Refractive index of alcohol w.r.t air, ^airμ_alcohol = 1.36;

Refractive index of flint glass w.r.t alcohol = ^alcoholμ_glass = ^airμ_glass / ^airμ_alcohol =

⇒ ^alcoholμ_glass = 1.213

Hence the refractive index of flint glass w.r.t alcohol is 1.213.

Absolute refractive index of glass =

Refractive index of diamond w.r.t glass is

⇒ μdiamond = 1.6 × μglass

⇒ μdiamond = 1.6 × 1.5 (μglass = 1.5).

⇒ μdiamond = 2.4

Hence the absolute refractive index of diamond is 2.4. Therefore the option B is correct.

According to Snell’s law

But i+r=90° (Reflected and refracted rays are at an angle of 900).

⇒

⇒ μ =.

⇒ μ =

⇒

Critical angle = μ =

⇒

⇒ sin i_c = tan r

⇒ i_c = sin^-1 (Tan r)

Therefore, the critical angle i_c = sin^-1(tan r)

hence the option A is correct.

According to the question following is the ray diagram.

As the incident ray is parallel to the principal axis. Therefore in absence of concave lens it will form an image at focus of convex lens.

∴ v = 20cm = f

Distance of Image I from the concave lens B will be = u = 20 – d

Now, the image I will act as object for the concave lens.

Since image object (image I) is at focus of concave lens. Therefore, Image will be formed at infinity

⇒ v =

Focal length = -5cm.

By Lens Formula.

⇒

⇒

⇒

⇒ 20 – d = 5

⇒ d = 20 – 5 = 15cm.

Therefore, the distance d will be 15cm. Hence the option B is correct.

Given :

Radius of curvature, R=-5 cm

Object distance, u=-3 cm

The refractive indices n₁=1.5 (glass), n₂=1.0(air)

We have , (The Lensmaker’s formula)

v= -2.5 cm

Then the image will appear at distance of 2.5 cm from the plane surface of the mirror

Hence D is the correct option.

If h₁ and h₂ are the sizes of the images formed in two conjugate positions, then, the size of object is given as h = √h₁h₂ = √9×1 = 3cm.

The size of the image in the two conjugate positions is

In the first case, if the image is formed, the magnification is given as . Such that, v = 3u

Also, v = 20+u, since “v” and “u” interchange the conjugate positions.

∴ 3u = 20 + u

Or 2u = 20

Or u = 10

And, v = 20+ u = 20+10 = 30 cm

Now, focal length is calculated as :

Thus, correct answer is option (d)