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As the scattered light is polarized, the sky appears bright and dim alternately.

Correct option is (C) 25 × 10^-7 m 

The wave length of light or refractive index of medium between objective lens and the object. 

Minimum separation OR Limit of resolution: 

d_min = 1.22λ / 2nsinβ 

where λ – wavelength of light, n – refractive index of the medium between the object and the objective lens and 2β – angle subtended by the object at the diameter of the objective lens at the focus of the microscope.

Resolving power of a microscope can be increased 

(i) by choosing a medium of higher refractive index and 

(ii) by using light of shorter wavelength.

Resolving power of the telescope = \(\cfrac{D}{1.22λ}\)

= \(\cfrac{1.22}{1.22\times5\times10^{-7}}\)

= \(\cfrac{10\times10^6}5\) = 2 × 10⁶ rad^-1

(A) zero or an integral multiple of λ

Expression for limit of resolution of telescope: 

Limit of resolution:   

Δθ = 0.61λ/a = 1.22λ/2a 

where λ – wavelength of light and 2a – diameter of aperture of the objective. 

Correct option is (B) \(\cfrac{1.22λ}D\)

Data : λ = 5500A = 5.5 × 10^-7 m, 

D = 2.44 m

Angular resolution. ∆θ = 1.22λ/D, D being the diameter of the aperture.

∴ ∆θ = 1.22 × 5.5 × \(\cfrac{10^{-7}}{2.44}\) = 2.75 × 10^-7 rad

= 0.0567 arcsec

In optical instruments, the sizes of the aperture is very large as compared to the wavelength of the light used. As such, the diffraction of light wave is of no significance.

(B) unequal width with varying intensity

The wavelength of sound waves is comparable with the size of the obstacles whereas for light, the wavelength is much smaller than the dimensions of most of the obstacles.  

No. As the order of the secondary maximum increases its width decreases. 

All the waves exhibit the phenomenon of diffraction. 

 Sound wave have frequencies 20 Hz to 20 kHz. The corresponding wavelengths are 15m and 15mm, respectively. Diffraction effects are seen if there are slits of width α such that.

α : λ

For light waves, wavelengths  10^–7m. Thus diffraction effects will show when

α : 10^–7 m.

whereas for sound they will show for

15mm < α < 15m .

(i) Obeying road rules, alertness, concern for other's life and clarity of knowledge. 

(ii) Due to the superposition of incident and reflected waves of white light by thin film (interference).

Diffraction Condition : The size of the obstacle sharpness should be comparable to the wavelength of the light falling.

Application : The finite resolution of our eye.

The fringe width is not affected by the intensity of incident light.

The fringe width is directly proportional to the wavelength of incident light. 

Correct option is (A) 1.5 W

Importance of Young’s experiment observe the interference of light: 

1. It was the first experiment (1800-04) in which the interference of light was observed. 

2. This experiment showed that light is propagated in the form of waves. 

3. From this experiment, the wavelength of monochromatic light can be determined.

With white light, one gets a white central fringe at the point of zero path difference along with a few coloured fringes on both the sides, the colours soon fade off to white.

The central fringe is white because waves of all wavelengths constructively interfere here. For a path difference of \(\cfrac12\)λ_violet, complete destructive interference occurs only for the violet colour; for waves of other wavelengths, there is only partial destructive interference. Consequently, we have a line devoid of violet colour and thus reddish in appearance. A point for which the path difference = \(\cfrac12\)λ_red is similarly devoid of red colour, and appears violettish. Thus, following the white central fringe we have coloured fringes, from reddish to violettish. Beyond this, the fringes disappear because there are so many wavelengths in the visible region which constructively interfere that we observe practically uniform white illumination.

1. Light is a transverse, electromagnetic wave. 

2. A light wave consists of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of propagation of the wave. 

3. Like all other electromagnetic waves, light waves do not require any material medium as they can travel even through vacuum. 

4. In a material medium, the speed of light depends on the refractive index of the medium, which, in turn depends on the permeability and permittivity of the medium.

(A) its wavelength decreases

The various theories of the nature of light from the 17th century to modern times are

1. Descartes’ corpuscular theory of light (1637) 

2. Newton’s corpuscular theory (1666) 

3. Huygens’ wave theory of light propagation (1678), modified, verified, and put on a firm mathematical base by Young, Fraunhofer, Fresnel, and Kirchhoff (in the 1800s) 

4. Maxwell’s electromagnetic theory (1865)

5. the light quantum, i.e., the photon model of the modern quantum theory by Planck (1900) and Einstein (1905).

Drawbacks of Newton’s corpuscular theory of light:

1. The theory predicted that the speed of light in a denser medium should be greater than that in air. This was disproved when experiment showed that the speed of light in water is less than that in air (carried out in 1850 by French physicist Jean Bernard Leon Foucault). , 

2. The theory could not satisfactorily explain the phenomenon of polarization and the simultaneousness of reflection and refraction. 

3. The corpuscular theory failed to explain the phenomena of diffraction and interference.

4. There was no basis for the hypothesis that the constituent colours of white light are due to different sized corpuscles.

Sir Isaac Newton developed the corpuscular theory of light proposed by Rene’ Descartes (1596-1650), French philosopher and mathematician. The theory assumed that light consists of a stream of corpuscles emitted by a luminous source.

Postulates of Newton’s corpuscular theory of light:

1. Light corpuscles are minute, light and perfectly elastic particles. 

2. A luminous source emits light corpuscles in all directions which then travel at high speed in straight lines in a given medium.

3. The constituent colours of white light are due to different sizes of the corpuscles. 

4. The light corpuscles stimulate the sense of sight on their impact on the retina of the eye. 

5. A reflective surface exerts a force of repulsion normal to the surface on the light corpuscles when they strike the surface. 

6. A transparent medium exerts a force of attraction normal to the surface on the light corpuscles striking the surface. This force is different for different mediums.

Notes :

1. A consequence of the assumption (6) is that, according to the corpuscular theory, the speed of light in a denser medium is greater than that in air and has different values for different mediums.

2. It was known from earliest recorded times that when light is incident on the surface of glass or water, it is partly reflected and partly transmitted, simultaneously. To explain this, Newton postulated that the corpuscles must have fits of easy reflection and fits of easy transmission and must pass periodically from one state to the other.

Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wavelets emanating from the wavefront are called as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time. 

Resultant intensity, I₀ 

= 2I₀ (1 + cos φ)

= 2I₀ (1 + cos 60°) 

= 2I₀ (1 + \(\cfrac12\)) = 3I₀.

Correct option is (B) wavefront

Plane polarised light is one which contains transverse linear vibrations in only one direction perpendicular to the direction of propagation.  

When an unpolarised light wave is incident on a polaroid, the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules. This direction is known as the pass-axis of the polaroid. 

Zero. Since no light passes through the polaroids when they are crossed.  

n = tan i_B, where n - refractive index of the reflector and i_B - polarising angle/Brewster’s angle. 

Brewster’s Law: The refractive index of a reflector is equal to tangent of the polarising angle. 

Polarisation, because polarisation is possible only for transverse wave. So all other phenomenon are due to super position of waves.

Using objective of larger diameter. 

(C) 2cos^-1 \(\left(\cfrac n2\right)\)

Data : α = 20°, 

λ = 6600 Å = 6.6 × 10^-7 m, 

n = 1 (air) resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope. 

Formula : Resolving power of a telescope

R = \(\cfrac 1θ\) = \(\cfrac D{1.22λ}\)

where θ ≡ the minimum angular separation of two closely-spaced celestial objects or the angular limit of resolution, D ≡ the diameter of the objective lens of the telescope, λ ≡ the wave length of light.

Advantages of a large objective lens in an astronomical telescope :

1. The resolving power is directly proportional to the diameter of the objective lens. Hence, a large objective lens results in a smaller Airy disc and a sharper image. 

2. It collects more of the incident radiation from a distant object which results in a brighter image.

Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle i.e

tan θ_B=n₂/n₁ where n₂<n₁.

When light travels in such a medium the critical angle is sin θ_c=n₂/n₁  

where n₂ < n₁.

As |tan θ_B|>|sin θ_c|for large angles, θ_B <θ_C .

Thus, polarisation by reflection shall definitely occur.

Data : λ = 650 nm = 6.5 × 10^-7 m, 

n = 1, 2α = 60°

∴ α = 30°

Resolving power of the microscope = \(\cfrac{2n \,sin \,α}λ\)

= \(\cfrac{2\times1\times\,sin\,30^\circ}{6.5\times10^{-7}}\) = \(\cfrac{10}{6.5}\) x 10⁶

In Young’s double-slit experiment, when white light is incident on the slits and one of the slit is covered with a red filter, the light passing through this slit will emerge as the light having red colour. The other slit which is covered with a violet filter, will give light having violet colour as emergent light. The interference fringes will involve mixing of red and violet colours. At some points, fringes will be red if constructive interference occurs for red colour and destructive interference occurs for violet colour. At some points, fringes will be violet if constructive interference occurs for violet colour and destructive interference occurs for red colour. The central fringe will be bright with mixing of red and violet colours.

(B) to increase the numerical aperture of the objective

Higher angular magnification of a high magnifying power microscope is of little use if the finer details in a tiny object are obscured by diffraction effects. Hence, a microscope of high magnifying power must also have a high resolving power.

Resolving power ot a microscope = \(\cfrac{2n\,sin\,α}λ\)

Where α ≡ the half angle of the angular separation between the objects, at the objective lens. n ≡ the refractive index of the medium between the object and the objective, λ ≡ the wavelength of the light used to illuminate the object.

The factor n sin α is called the numerical aperture of the objective and the resolving power increases with increase in the numerical aperture. To increase α the diameter of the objective would have to be increased. But this increase in aperture would degrade the image by decreasing the resolving power. Hence, in microscopes of high magnifying power, the object is immersed in oil that is in contact with the objective. Usually cedarwood oil having a refractive index 1.5 (close to that of the objective glass) is used. Closeness of the refractive indices also reduces loss of light by reflection at the objective lens.

Solve for the speed (u_g) of light in glass and the angle of refraction (r) in glass as in Solved Problem (14). 

u_g = 2 × 10⁸ m/s and r = 35°16′ 

The angle of incidence (z), i.e., the angle between the incident ray and the normal to the glass surface, is i = 90° – 30° = 60°. 

Hence, the angle by which the refracted ray is deviated from the original path is δ = i – r = 60° – 35°16′ = 24°44′

When a beam of unpolarized light is passed through certain types of materials (or devices), these materials (or devices) allow only those light waves to pass through which have their electric field along a particular direction. All the other waves with the electric field in other directions are blocked. A material (or a device) which exhibits this special property is called a polarizer.

According to the electromagnetic theory of light, a light wave consists of electric and magnetic fields vibrating at right angles to each other and to the direction of propagation of the wave. If the vibrations of the electric field \(\vec E\) in a light wave are confined to a single plane containing the direction of propagation of the wave so that its electric field is restricted along one particular direction at right angles to the direction of propagation of the wave, the wave is said to be plane-polarized or linearly polarized.

If the vibrations of \(\vec E\) in a light wave are in all directions perpendicular to the direction of propagation of the light wave, the light wave is said to be unpolarized. Ordinary light, e.g. that emitted by a bulb, is unpolarized.

According to Biot, unpolarized light may be considered as a superposition of many linearly polarized waves, with random orientations. Also, these component waves are noncoherent, that is, irregular in their phase relationships.

[Note : Ordinary light consists of wave trains, each coming from a separate atom in the source. A beam of ordinary light in a single direction consists of millions of such wave trains from the very large number of atoms in the source radiating in that direction. Hence, the vibrations of \(\vec E\) are in all transverse directions with equal probability. Thus, light from an ordinary source is.un-polarized.]

Correct option is: (D) polarization

Data : a = 0.5 cm = 0.5 × 10^-2 m, 

D ≃ f = 40 cm = 0.4 m, λ = 4890 Å = 4.890 × 10^-7 m

The distance between the first dark fringe and the next bright fringe = \(\cfrac{λD}{2a}\)

= \(\cfrac{4.890\times10^{-7}\times0.4}{2\times0.5\times10^{-2}}\) = 1.956 x 10^-5 m

Characteristics of a single-slit diffraction pattern :

1. The image cast by a single-slit is not the expected purely geometrical image.

2. For a given wavelength, the width of the diffraction pattern is inversely proportional to the slit width. 

3. For a given slit width a, the width of the diffraction pattern is proportional to the wavelength. 

4. The intensities of the non-central, i.e., secondary, maxima are much less than the intensity of the central maximum. 

5. The minima and the non-central maxima are of the same width, Dλ/a. 

6. The width of the central maximum is 2Dλ/a. It is twice the width of the noncentral maxima or minima.