Light - Reflection And Refraction

Light rays passing through the focus of a concave mirror can makes a parallel beam of light. Similarly light rays passing through a convex lens can make parallel beam of light.

The magnification is given by:

On calculating, we get u= -60 cm

Now, on applying the mirror formula, we get

Thus, the focal length is, F = -20 cm

When the object is placed between the focus and the center of curvature of a concave mirror, the image formed will be beyond C, enlarged in size, Real and inverted.

It can be show with the following ray diagram:

According to Snell’s law:

As it is clear from the figure, when the light rays travelled from medium A to medium B, then they bend towards the normal which means that medium B has higher refractive index and less speed of light with respect to medium A. So, refractive index of medium B with respect to medium A will be greater than unity.

Since the emergent rays are parallel to the incident rays direction and we know that out of convex lens, concave lens, prism and rectangular glass slab, only glass slab can change the light ray path so that emergent rays are parallel to incident rays. Therefore, correct answer is a rectangular glass slab.

Since a convex lens can converge the light rays at a point and emerged at point B. So, the convex lens is inside the box.

Since power of a lens is given by the relation:

P = 1/f

So, a convex lens has 4 diopter power having a focal length of 0.25 m.

Since, Magnification is defined as the ratio of height of the image to the height of object and we know that a rear view- mirror i.e. convex mirror forms a virtual erect and diminished image. So, Magnification produced by convex mirror will be less than one.

We know that in case of a concave mirror, when the object is placed at infinity, then the light rays after reflection gets converged at focus of the concave mirror. Sun is at the infinity point, So focal length of the mirror will be 15 cm. The object should be placed between at 2f to obtain the same size of image as that of the object.

So, object distance = 2f = 2 x 15 = 30 cm

As we know that, a convex mirror forms virtual, erect and diminished sized images. So, a full length image of a distant tall building can be achieved by taking its diminished image by convex mirror.

Since light rays when falls on a mirror, they become parallel to the principal axis and we get a concentrated light ray. This is the reason why the bulb is placed very near to the focus of the reflector or mirror.

The laws of reflection hold good for all the mirrors having smooth surface for light reflection irrespective of their shape.

Since when path of a ray of light coming from air when passes through a rectangular glass slab, then the emergent ray is parallel to the incident ray direction which is correctly shown in figure B.

Since Glycerin has highest refractive index among kerosene, Water and mustard oil, so the ray of the light incident obliquely at the some angle would bend the most in Glycerin medium.

The light rays parallel to the principal axis will pass through the focus of the mirror after reflection of light. So, figure D is correct.

When a ray of light passes through the focus of a convex lens, then after the reflection of light, the ray becomes parallel to the principal axis of the convex lens. So figure A is correct.

Since the image of her head is big, it means that the concave mirror has been used of higher focal length. The middle portion of the body of the child remains of same size which confirms the usage of plane mirror in the middle and in the bottom, the legs appears smaller which means that convex lens has been used in the bottom. So, the correct combination is: concave, plane mirror and convex mirror.

When the object is placed at infinity, then the incident rays fall parallel on the principal axis, which on after reflection by a concave mirror, convex mirror and convex lens passes through the focus of the reflector. The image thus obtained is diminished and point sized.

(a) The device when Object is placed between device and its focus, image formed is enlarged and behind it. Will be a concave mirror.

(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object, then the device is a convex lens. It can be shown with the following ray diagram:

(c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object, then the device will be a concave lens. The ray diagram is:

(d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it, then the device used will be a convex mirror. The ray diagram will be as follows:

When light rays is incident on the rectangular glass slab immersed in any medium , then it is refracted by the glass slab and after refracting it comes out of the glass and gets another interface of glass and air, so on going from denser to rarer, it bends away from the normal ray. As the glass slab is rectangular in shape and both sides have the same medium. So, the light ray which is incident and the emergent rays are parallel. It can be shown with the help of following diagram:

The pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water due to the refraction of light but if we use liquid like kerosene in place of water, then the pencil will bend more than it bend in case of water because the refractive index of kerosene is higher than that of water and is therefore optical denser than water.

So, the pencil will not bend to the same extent due to different refractive indices of different media.

The refractive index of a medium is related to the speed of light by the following relation:

n = c/ v where n is the refractive index

c is the speed of light

v is the velocity of the light in the medium

The expression of refractive index of a medium w.r.t another medium in terms of speed of light in these two media is as follows:

Let light travels from a medium say air to a medium 1, then the refractive index of medium 1 can be written as:

……..(1)

When light rays are travelling from air to a medium 2, then the refractive index of medium 2 can be written as:

……….(2)

Dividing equation (2) by equation (1), we get

Refractive index of diamond with respect to glass n_dg = 1.6

Absolute refractive index of glass = 1.5

Now

Putting the values 1.6 =

n_d =

Yes, the statement is correct.

To get the magnified virtual image, the object is to be placed between the optical centre of the convex lens and the focus F₁ of the convex lens. The image formed will be virtual and magnified.

However, for obtaining a magnified real image, the object should be placed in between F₁ and 2F₁ of the convex lens.

To get a sharp image of the object, Sudha should move the screen towards the lens and keeping the lens fixed because the object is now placed at beyond 2f and the lens will form the image towards on opposite side between F and 2F.

Since the actual distance between the window pane and the mirror is 15 cm. So, the approximate focal length of the lens will be approximately 15 cm.

Power and focal length of a lens are related to each other by the following relation:

P = 1/ f

Where P is the power of lens

f is the focal length of the lens and should be in m.

The lens having high power or small focal length will have more converging power. So, the lens to obtain more convergent light is of focal length 20 cm.

The incident ray and reflected ray will always be parallel to each other when the two plane mirrors will be placed at right angle i.e. 90⁰ to each other.

The diagram is as follows:

(i) Ray diagram when the path of rays of light enters from air to water:

(ii) Ray diagram when the path of light rays enters from water into air:

(a) Ray diagram when object is placed between the pole and focus of the concave mirror:

(b) Ray diagram when the object is placed between the focus and centre of curvature of concave mirror:

(c) Ray diagram when the object is placed at the centre of curvature of the concave mirror:

(d) Ray diagram when the object is placed at little beyond centre of curvature of the concave mirror:

(e) Ray diagram when the object is placed at infinity of the concave mirror:

(a) Ray diagram when an object is placed between optical centre and focus of the lens:

(b) Ray diagram when an object is placed between focus and twice the focal length of the lens:

(c) Ray diagram when an object is placed at twice the focal length of the lens

(d) Ray diagram when an object is placed at infinity

(e) Ray diagram when the object is placed at the focus of the lens:

The laws of refraction are:

(a) The two laws of refraction of light are as follows:

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

2. The second law of refraction is also known as Snell’s law of refraction and it states that: the ratio of Sine of the angle of incidence to the sine of refraction is constant for a given pair of media. It establishes a relation between angle of incidence and angle of refraction.

It can be expressed mathematically as follows:

It can be shown with the following diagram:

Here, EO is the incident ray, OO’ is the refracted ray and O’H is the emergent ray.

Since the incident ray, refracted ray and the emergent ray are in the same plane which is the first law of refraction. Moreover as light ray enter from a rarer to a denser medium, it bends towards the normal and when it comes out from glass to air, it bends away from the normal. The refractive index can be calculated by taking the ratio of sine of the angle of incidence to the ratio of sine of refraction which is also the Snell’s law.

(a) The ray diagram when the object is placed at the focus of the concave lens:

(b) The ray diagram when the object is placed between focus and twice the length of focal length of the lens:

(c) Ray diagram when the object is beyond twice the focal length of the concave lens:

(a) Ray diagram when the object is placed at infinity of the convex mirror:

(b) Ray diagram when the object is placed at finite distance from the convex mirror:

Given: Magnification = 3

Distance of image from the lens = -80 cm

Now, Magnification m is given by

u=26.67 cm

So, the object is to be placed at 26.67 cm from the lens.

The nature of the image is virtual, erect and larger than the object size.

Applying the lens formula:

F = 20 cm

Since focal length of the lens is positive, so the lens is convex lens.

Focal length of the mirror f =20 cm

h_i/h₀ = 1/3

if the mirror is a concave mirror, then the focal length of the mirror, f= -20 cm

Magnification is given by

Object distance u = -3v or v= - u/3

Applying the mirror formula

v = 40/3

So, the image is virtual, erect and diminished and the mirror is a concave mirror.

Power is defined as the degree of divergence of convergence achieved by the lens and it is reciprocal of the focal length.

The unit of Power is Dioptre.

Focal length = 50 cm

In meter, focal length is : 0.5 m

Power = 1/f = 1/0.5

Power of the lens = +2D

The lens will be a convex lens

Focal length = -50 cm

In meter, focal length will be = -0.5 m

Power = 1/-0.5 = -2D

In this case, the lens will be a concave lens.

(i)object distance u = position of convex lens – position of candle

u = - (50 -12)= -38 cm

image distance v = position of screen –position of convex lens

v= 88-50 cm

v= 38 cm

Applying the lens formula:

on calculating focal length willbe f = 19 cm

(ii) When the candle is shifted towards the lens at a position of 31.0 cm:

Object distance u =-(50-31)

u = -19 cm

Applying the lens formula

Therefore, v is infinity

(iii) If the candle is further shifted towards the lens, then the nature of the image formed will be virtual, erect and magnified

(iv) The ray diagram will be as follows: