Reflection and Refraction Test - 1

CONCEPT:

• Paraxial rays: A ray which makes a small angle (θ) to the optical axis of the system and lies close to the axis throughout the system is called paraxial rays.
• Marginal rays: The rays which pass through the maximum aperture of the spherical mirror is called marginal rays.

The lens maker’s formula is given by:

\(\frac{1}{f} = \left( {n - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\)

Where, f is the focal length (half the radius of curvature), n is the refractive index of the material used, R₁ is the radius of curvature of sphere 1 and R₂ is the radius of curvature of sphere 2

EXPLANATION:

• The above formula is only valid when the lens is thin. If the lens will be thick then the formula will be complicated and different.
• When the rays are paraxial then the image formed will be more accurate and will be a point image and the above formula will give the exact location.
• Thus the rays must be paraxial and the lens must be thin to validate the lens maker’s formula. So option 2 is correct.

CONCEPT:

• Converging lens: A lens in which light rays enter into it parallel to its axis and converge at a single point on the opposite side.

• Focal length: When light passes through a lens, the measure of how strongly the system converges or diverges the light.
  • The focal length is the inverse of the system's optical power.
• Magnification: In a convex lens, the magnification is the ratio of the height of the image to the height of the object.

For near point D, magnification of the convex lens is given by:

 \(m = 1+\frac{D}{f}\)

where m is magnificent, D is the near point, and f is the focal length of the convex lens.

CALCULATION:

Given that D = 25 cm and f = 5 cm.

magnification \(m = 1+\frac{D}{f}\)

\(m = 1+\frac{25}{5}\)

m = 6 

So the correct answer is option 4.

CONCEPT:

• Refraction of Light: The bending of the ray of light passing from one medium to the other medium is called refraction.

• The refraction of light takes place on going from one medium to another because the speed of light is different in the two media.
• The greater the difference in the speeds of light in the two media, the greater will be the amount of refraction.
• A medium in which the speed of light is more is known as optically rarer medium and a medium in which the speed of light is less is known as an optically denser medium.

EXPLANATION:

• A lemon kept in water in a glass tumbler appears to be bigger than its actual size when viewed from the sides. This is because of the refraction of light. Therefore option 3 is correct.
• Light is refracted as it passes from water into the air and as we know the refracted ray bends away from normal when it passes from denser to rarer medium.
• Water in a glass tumbler has a round outer surface that is convex and is a magnifying glass.
• As light is bent and being bent to the point that the object being observed appears to be larger than its normal size.

Concept:

• Power of a lens is its ability to bend light and it is equal to the reciprocal of the focal length of the lens in meters.        
• P  = 1/f         where f is focal length of the lens in meters
• The equivalent power for a system, where two thin lenses are sharing an axis are kept in contact with each other, is given by the following formula:

P = P₁ + P₂

Where,    P₁ is the power of first lens and P₂ is the power of second lens.

Calculations:

Given;  P₁ = P₂ = 2D

Equivalent power = P = P₁ + P₂ = 2 + 2 = 4D

Focal length of the combination = f = 1/P = ¼ = 0.25 m

Points to remember: Unit of Power is taken as Diopter (D) only when focal length is taken in meters. If focal length is in cm then power is equal to cm^-1.

CONCEPT:

• Lens: The transparent curved surface which is used to refract the light and make an image of any object placed in front of it is called a lens.
• Concave lens: It is thin in middle and thick at the edges.
  • Actually, a concave lens is a spherical lens whose two surfaces are concave or bent inward.
  • A concave lens is also know known as the ​diverging lens.
• Lens formula is given by:

\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

• The magnification of the lens is given by:

Magnification (m) = v/u

Where u is object distance, v is image distance and f is the focal length of the lens

CALCULATION:

Given - focal length (f) = -20 cm and object distance (u) = -40 cm

• Lens formula is given by:

\(⇒ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

\(⇒ \frac{1}{v} = \frac{1}{f}+\frac{1}{u}\)

\(⇒ \frac{1}{v} =\frac{1}{{ - 20}} + \left( {\frac{1}{{ - 40}}} \right)=\frac{-1}{20}-\frac{1}{40}=\frac{-2-1}{40}=\frac{-3}{40}\)

⇒ v = -13.33 cm

Concept:

Image Formation by Convex mirror

• At any finite position of an object, the image formed by a convex mirror is on focal length. 
• The image is formed by virtual intersection light and hence is always virtual.

Mirror Formula: 

• The following formula is known as the mirror formula, which represents the relation between focal length, image distance, and object distance
• \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

  where, f = focal length, v = the distance of the image from the mirror, and u = the distance of the object from the mirror.

• Sign convention is followed in mirror formula

Sign convention in mirrors

• All distance are measured from the pole of the mirror.
• The distance measured from the pole (P) to the left side of the mirror is taken as negative.
• The distance measured from the pole (P) to the right side of the mirror is taken as positive.
• The heights measured upwards and perpendicular to the principal axis of the mirror is taken as positive.
• The heights measured downwards and perpendicular to the principal axis of the mirror is taken as negative.
• In a convex mirror, the focus is behind the mirror so, the focal length is positive.

Calculation:

Object distance u = - 30 cm  (following sign convention)

Focal length f =  30 cm 

Image distance v = ?

Now applying the mirror formula, we get 

\(\frac{1}{30}=\frac{1}{-30}+\frac{1}{v}\)

\(\implies \frac{1}{v}=\frac{1}{30}+\frac{1}{30}\)

\(\implies \frac{1}{v}=\frac{2}{30}\)

⇒ v = 15 cm

The sign obtained is positive and the image distance is 15 cm.

So, By the sign convention, the image is formed 15 cm behind the mirror. 

"15 cm behind the mirror" is the correct option.

CONCEPT:

• Rayleigh's law of scattering: According to Rayleigh's law of scattering, the intensity of light of wavelength λ present in the scattered light is inversely proportional to the fourth power of λ, provided the size of the scattering particles is much smaller than λ.

Mathematically,

\(I \propto \frac{1}{{{\lambda}^4 }}\)

• Thus the scattered intensity is maximum for a shorter wavelength.

EXPLANATION:

• The sun looks reddish at the time of sunrise and sunset:

• At the time of sunrise and sunset, the sun is near the horizon. The rays from the sun have to travel a larger part of the atmosphere.
•  As the wavelength of red colour is more than that of blue colour (λb≪< λr) and the intensity of scattered light is \(\propto\frac{1}{{{\lambda ^4}}}\), therefore, most of the blue light is scattered away.
• Only a red colour, which is least scattered enters our eyes and appears to come from the sun. Hence the sun looks red, both at the time of sunrise and sunset.
•  The red color of the sun at sunset and sunrise is due to the scattering of sunlight. 

Additional Information

CONCEPT:

• Optical fibers are transparent fibers and act as a light pipe to transmit light between its two ends. They are made up of silicon dioxide.

• The total internal reflection occurs when the angle of incidence is greater than the critical angle.
• In optical fiber glasses of a high and lower index are assembled in the precise order.
• The light comes in from one end of the fiber and after thousands of successive internal reflections, the light reaches the opposite end of the fiber with almost zero loss. 

EXPLANATION:

• The refractive index of the material of the core (μ1) is higher than that of the cladding (μ2). Therefore option 1 is correct.
• When the light is incident on one end of the fiber at a small angle, the light passes inside, undergoes repeated total internal reflections along with the fiber, and finally comes out.
• Hence, the function of the cladding is to provide a lower refractive index at the core interface in order to cause reflection within the core so that light waves are transmitted through the fiber.
• Thus, cladding is used in the optical fiber in order to cause Total internal reflection.

CONCEPT:

Total internal reflection (TIR):

• Total internal reflection (TIR) occurs at the interface of two transparent medium when the ray of light travels from a denser medium to a rarer medium.
• The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.
• Total internal reflection (TIR) of light is the reflection of light within the same medium when a ray of light is incident at the interface of two medium at an angle of incidence which is greater than the critical angle for the pair of media i.e. i > ic.
• The relation between refractive index and critical angle is

\(μ =\frac{1}{\sin C}\)

CALCULATION:

Given  - critical angle (C) = 30∘

• The relation between refractive index and critical angle is

\(\Rightarrow \mu =\frac{1}{\sin C}\)

\(\Rightarrow \mu =\frac{1}{\sin 30}=2\)

As we know, the refractive index is written as

\(\Rightarrow \mu=\frac{c}{v}\)

\(\Rightarrow v = \frac{3\times 10^8}{2}=1.5\times 10^8\, m/s\)

CONCEPT:

• Plane Mirror: A plane mirror is a mirror with a flat (planar) reflective surface.

The characteristics of an image formed in a plane mirror:

• The image formed by the plane mirror is virtual and erect i.e. image cannot be projected or focused on a screen.
• The distance of the image ‘behind’ the mirror is the same as the distance of the object in front of the mirror.
• The size of the image formed is the same as the size of the object.
• The image is laterally inverted, i.e. left hand appears to be right hand when seen from the plane mirror.
• If the object moves towards (or away from) the mirror at a certain rate, the image also moves towards (or away from) the mirror at the same rate.

EXPLANATION:

Given that:

OP = 0.5 m and OI = ?

As we know, the distance of the image ‘behind’ the mirror is the same as the distance of the object in front of the mirror. Therefore, PI = 0.5 m

Hence, OI = OP + PI

OI = 0.5 m + 0.5 m = 1 m