Light-reflection And Refraction

The magnification produced by a plane mirror is +1. Because,

And the height of image formed by plane mirror is equal to height of object.

Therefore, Height of Image = Height of Object.

Hence Magnification is +1.

When the object is kept at the centre of curvature of the concave mirror the image of object coincide with the object, because image is formed at the centre of curvature and is same as the size of the object as shown in the figure below.

Due to higher refractive index of water (1.33) than air (1.0003) the air bubble acts as the diverging lens (Concave lens).The figure below illustrates it.

A concave mirror is used in the headlight of a car because concave mirror is a converging mirror. In headlights the bulb is kept at the focus of the mirror, due to which the rays emerging from the bulb after reflection from the mirror becomes parallel to the principal axis hence we get the powerful beam of light which can go to farther distances.

₁n₂ and

₂n₁

⇒ ₁n₂ × ₂n₁ = = 1.

Hence the answer is 1

Let velocity of light in vacuum=v

⇒ Velocity of light in water

Now,

Refractive index of a medium (water) =

⇒ Refractive index = =

Hence Refractive index of water = 1.33.

In plane mirror the distance between object and mirror is same as the distance between the image and mirror.

⇒ Image Distance = Object Distance.

∴ Distance between image and plane mirror is 15 cm.

Method We can put the object far away from the concave mirror and place the screen where the image is formed. The place where the image formed is sharpest is the focus of the mirror, measure the distance between the focus and mirror to get the focal length.

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

According to the question;

Focal length (f) = -1.5m;

Object distance (u) = -40cm = 0.40m; (1m = 100 cm)

By mirror formula;

⇒

⇒

⇒ v = 0.5454m = 54.54cm (1m = 100 cm).

Since v = 54.54cm which is positive hence image is behind the mirror.

The image is formed at the distance of 54.54 cm behind the mirror and is virtual and erect.

Person will use the convex mirror because it always forms virtual and erect image and has the wider field of view which will help in capturing the whole building.

The figure below shows the image of tall building in convex mirror.

This is due to the 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 cause the depth to appear higher and we cannot see the actual depth of the lake.

The figure below illustrates the same.

Refractive index of air, μ_air = 1;

Given;

Refractive index of water w.r.t air, ^aμ_w;

Refractive index of glass w.r.t air, ^aμ_g;

Refractive index of water w.r.t glass

⇒ ^gμ_w =

∴ Refractive index of water w.r.t glass is 8/9.

Also;

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

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

∴ Refractive index of glass w.r.t water =
Refractive index of glass w.r.t water = 9/8.

The focal length of a mirror does not depend upon the nature of the medium in which it is placed whereas the focal length of a lens depends upon the medium in which it is placed. Thus, there will be no change in the focal length of the concave mirror whereas the focal length of the convex lens will change. If the refractive index of the material of the convex lens is greater than water it will increase, if it is equal to water it will become infinity and if it is less than water it will be negative.

The lens is concave lens, hence it diverges the ray of the light coming from infinity after refraction;

(b)

The above lens is the convex lens; hence the ray of light passing through focus after refraction will become parallel to the principal axis after refraction.

According to the question;

Focal length (f) = 15cm;

Radius of curvature(R=2f) = 30cm;

Image distance (u) = 30cm;

Since the image distance is 30 cm and the Radius of curvature of the lens is also 30 cm.

∴ the object must be placed at the radius of curvature (30cm) in front of lens, because when the object is kept at radius of curvature its image is also formed at radius of curvature on other side of the lens.

Figure below shows the image formation.

Object should be placed at 30cm in front of lens.

Position of Image: Between Pole and Focus

Size of Image: Diminished

Nature of Image: Virtual and erect.

Ray diagram:

Position of Image: Beyond the centre of curvature.

Size of Image: Magnified

Nature of Image: Real and inverted

Ray diagram:

Refractive index of transparent medium is defined as the ratio of speed of light in vacuum to the speed of light through medium.

Refractive index (n) =

Where;

C = speed of light;

V = speed of light in medium;

It has no unit because it is the ratio of two quantities of same type (speed);

Glass has higher refractive index of 1.52. (Because Refractive index of water = 1.33).

According to the question;

Object distance (u) = -45cm;

Image distance (v) = 90cm;

By lens formula;

⇒ f = 30cm.

Since f = 30cm which is positive hence it is a convex lens of focal length 30cm.

Now;

Height of Flame h₁= 2cm;

Magnification = =

Putting values of v and u

Magnification =

⇒

⇒

Height of image is 4 cm.

Negative sign means image is inverted.

According to the question;

Object distance (u) = -30cm;

Focal length (f) = -15cm;

Image distance = v;

By mirror formula;

⇒

⇒ v = 30cm.

Thus, screen should be placed 30cm in front of the mirror (Centre of curvature) to obtain the real image.

Height of object h₁= 2cm;

Magnification = =

Putting values of v and u

Magnification =

⇒

⇒

Height of image is 2 cm.

Negative sign means image is inverted.

Thus real, inverted image of size same as that of object is formed.

Diagram below shows the image formation.

According to the question;

Object distance = u;

Image distance (v) = 20cm;

Focal length = 10cm

By lens formula;

⇒ u = -20cm.

Therefore, object is placed at 20 cm in front of lens.

Now;

Height of object h₁= 2cm;

Magnification = =

Putting values of v and u

Magnification =

⇒

⇒

Height of image is 2 cm.

Negative sign means image is inverted.

Thus, the image is real, inverted and same as size of object.

Diagram below shows the image formation.

According to the question;

Object distance = u;

Image distance (v) = -15cm;

Focal length = -20cm

By lens formula;

⇒ u = -60cm.

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

Now;

Height of object h₁= 5cm;

Magnification = =

Putting values of v and u

Magnification =

⇒

⇒

Height of image is 1.25 cm.

Positive sign means image is virtual.

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.

f₁ = 50cm = 0.5m (1m = 100cm)
⇒ P₁ = 1/f1
P₁ = 1/0.5
P₁ = 2D
Since Power is positive therefore the lens with focal length 50cm is convex lens.
f₂ = -50cm = -0.5m (1m = 100cm)
⇒ P₂ = 1/f1
P₂ = 1/-0.5
P₂ = -2D
Since Power is negative therefore the lens with focal length -50cm is concave lens.
The image formed by the concave lens is virtual, erect and diminished irrespective of the position of the object.

According to the question;

Object distance = -12cm;

Image distance = v;

Focal length = 18cm;

By lens formula;

⇒ v = -36cm.

Therefore, image is formed at 36 cm in front of lens.

Now;

Height of object h₁= 5cm;

Magnification = =

Putting values of v and u

Magnification =

⇒

⇒

Height of image is 15 cm.

Thus, the image is virtual, enlarged, and erect and three times the size of object.

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

P (in dioptre) =

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

1 D =

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

Given;

Focal Length of convex lens f₁ = 25cm = 0.25m (1m = 100cm);

Focal Length of concave lens f₂ = -10cm = -0.1m (1m = 100cm);

Power of convex lens P₁ =

∴ P₁ = 4D.

Power of concave lens P₂ =

∴ P₂ = -10D.

Power of combination = P₁ + P₂ = 4-10 = 6D.

Thus, power of combination is -6 Dioptre.

Refraction of light is the phenomenon of bending of light as it travels from one medium to the other. When a ray of light travels from one transparent medium to the other it bends at the surface, this is due to the different optical density of different medium.

The figure below illustrates the refraction of light.

Snell’s Law states that ratio of sine of angle of incidence and angle of refraction of two medium is always constant.

Mathematically it is given as

^aμ_b

^aμ_b = is the refractive index of medium a with respect to medium b

Given;

Refractive index of (R.I) water w.r.t vacuum;

Refractive index (R.I) of vacuum w.r.t glass ;

Refractive index (R.I) of glass to vacuum will be =

Refractive index of glass to water =

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

⇒

⇒

⇒ speed of light in water

⇒ speed of light in water

Thus, speed of light in water = 2.25 × 10⁸.

Refractive index of water w.r.t vacuum = ^vμ_w;

⇒

⇒

⇒

Thus, speed of light in vacuum is 3 × 10⁸ m/s.

Concave lens works on the principle of refraction. We can divide the lens into tiny pieces which resembles a triangular prism (shown in figure), thus when the light ray fall it refracts the light either inwards or outwards due to its shape. Concave lens is thicker outside and thinner at edges. So the light rays are refracted outside the figure below illustrates it more clearly.

According to the question;

Object distance = u;

Image distance (v) = -10cm;

Focal length = -15cm;

By lens formula;

⇒ u = -30cm.

Thus, object should be placed 30cm in front of lens.

Now;

Height of object h₁= 2cm;

Magnification = =

Putting values of v and u

Magnification =

⇒

⇒

Height of image is 1.67 cm.

Thus, the image is virtual, diminished, and erect and one-third of the size of object.

Position of Image: At the centre of curvature.

Nature of Image: Real and inverted and same as object size

Ray diagram:

Position of Image: Behind the mirror

Nature of Image: Virtual, erect and magnified.

Ray diagram:

Position of Image: Behind the mirror

Nature of Image: Virtual, erect and diminished.

Ray diagram:

Position of Image: At 2f Behind the lens.

Nature of Image: Real, inverted and same as object size.

Ray diagram:

Position of Image: Behind the lens

Nature of Image: Virtual, erect and diminished.

Ray diagram: