Wave Optics

Given, path difference and
Wavelength



Here, is odd multiple of , so point is dark.

The optical path between any two points is proportional to the time of travel.
The distance traversed by light in a medium of refractive index in time t is given by

where v is velocity of light in the medium. The distance traversed by light in a vacuum in this time,



This distance is the equivalent distance in vacuum and is called optical path.
Here, optical path for first ray =
Optical path for second ray =
Path difference =
Now, phase difference
path difference

When the point source or linear source of light is at very large distance, a small portion of spherical or cylindrical wavefront appears to be plane, such a wavefront is called a plane wavefront.

In the given options none of sources generates plane wavefront, it can be artificially produced by reflection from a mirror or by refraction through a lens.

Fresnel used a biprism to obtain two coherent sources for producing interference fringes in the laboratory. It consists of two acute angled prisms with their bases in contact. In actual practice, the biprism is constructed as a single prisms of obtuse angle of about 179 and the remaining two acute angles are 30' each. In bi prism, the virtual images act as two coherent superimposed and interference fringes are formed in overlapping region AB on a screen placed at 0.

In order to measure the distanced between the virtual sources and , Glazebrook gave a method, known as magnification method due to Glazebrook. If and are distance between real images of and , then
Given

Photoelectric effect states that light travels in the form of bundles or packets of energy, called photons. This effect is explained on the basis of quantum nature of light. So, it clearly explains the particle's nature of light.

Here,

...(i)
Moreover, ...(ii)
From Eqs. (i) and (ii), we get

Concept: Physics (Optics) - Young's Double Slit Experiment (YDSE).

Step 1: State the formula for the distance of the maximum. The distance of the bright fringe (maximum) from the central maximum is given by: $ n \lambda D d_{slit} D d_{slit} d \propto n\lambda 8^{th} \lambda_1 d_1 \propto 8\lambda_1 6^{th} \lambda_2 d_2 \propto 6\lambda_2 $

From Brewsters laws
i.e., the polarising angle depends on refractive index and hence on wavelength and therefore is different for different colours.

Distance between the first minimum on other side

Resolving limit of microscope is given by

where, is called numerical aperture [NA] and wavelength of light.

Hence, if numerical aperture of a microscope is increased, then its limit of resolution is decreased.

We know that,


On adding Eqs. (i) and (ii), we get



But
Therefore,

Step 1: Understanding the formula for resolving power.
In the formula for resolving power of a microscope, the numerical aperture is defined as: Where: - is the refractive index of the medium, - is the half-angle of the cone of light entering the microscope, - is the wavelength of light. Thus, the correct answer is (C) .

Step 1: Understanding polaroids.
When unpolarised light passes through the first polaroid, it becomes polarised. However, if the second polaroid is crossed with the first one (i.e., the transmission axis of the second polaroid is at 90° to the first), no light will pass through the second polaroid. Step 2: Conclusion.
Thus, the correct answer is (C) the light is blocked by the second polaroid.

Step 1: Resistance of the coil.
The resistance of the coil is .
Step 2: Effective resistance of the circuit.
The distance between and is , and the total resistance in the loop includes the resistance of the coil and the internal resistance of the battery, which is . The total resistance in the circuit is:
Step 3: Apply Ohm's law.
Using Ohm’s law, the current is:
Step 4: Conclusion.
The current flowing through the circuit is 2 A, so the correct answer is (D).

Step 1: Understanding interference pattern.
In interference, when two light waves with different amplitudes interfere, the result is not complete destructive interference. Instead, there is some residual intensity even in the region where the waves would normally cancel each other out if they had the same amplitude. This occurs because the waves do not cancel completely due to their different amplitudes.
Step 2: Conclusion.
Thus, the correct answer is (C) there is some intensity of light in the region of destructive interference.

Step 1: Recall Brewster’s law.
At the polarising (Brewster) angle, the reflected light becomes completely plane polarised.

Step 2: Behaviour of refracted (transmitted) light.
The refracted ray still contains vibrations in more than one direction, hence it is only partially polarised.

Step 3: Eliminate incorrect options.
- Complete polarisation of refracted light does not occur.
- Planes of polarisation are not both completely polarised.

Step 4: Conclusion.
Thus, reflected light is plane polarised while transmitted light is partially polarised.

Step 1: Photoelectric equation.

Step 2: Relation between frequency and wavelength.

Step 3: Comparison of wavelengths.
Blue light has shorter wavelength than yellow light, hence higher frequency and higher photon energy.
Step 4: Effect on stopping potential.
Higher photon energy increases maximum kinetic energy of photoelectrons, therefore stopping potential increases.
Step 5: Conclusion.
The stopping potential will be more than 4 V.

Step 1: Lens formula.
For a convex lens,

Step 2: Mean position of the object.
Given:



Step 3: Linear magnification.

Step 4: Amplitude of image.
Amplitude of object . Hence,

Step 5: Conclusion.
The amplitude of oscillating image is nearly 2 cm.

Step 1: Understanding diffraction.
For diffraction at the first maximum, the path difference between the rays coming from the edges of the slit is . The phase difference corresponding to a path difference of is radians.
Step 2: Conclusion.
Thus, the phase difference between the rays coming from the edges of the slit is radians, corresponding to option (B).

Step 1: Write the formula for fundamental frequency of a stretched string.

Step 2: Parameters of the second string.
Length , linear density , tension .

Step 3: Write the frequency of the second string.

Step 4: Simplify the expression.

Step 5: Express in terms of .

Step 6: Conclusion.
The fundamental frequency of the second string is .

Step 1: Pressure at depth .

Step 2: Substitute values.

Step 3: Use compressibility relation.
Fractional compression is given by:

Step 4: Substitute values.

Step 5: Conclusion.
The fractional compression of water is .

Step 1: Critical Angle Formula.
The critical angle is given by the formula: For a medium with polarisation angle , the refractive index is related to the tangent of the angle. Thus, the critical angle is given by: Step 2: Conclusion.
Thus, the critical angle is .

Step 1: Fringe Separation Formula.
The fringe separation in a double slit experiment is given by: where is the wavelength of the light, is the distance between the slits and the screen, and is the separation between the slits. The position of the bright fringe is given by: For the first light with wavelength , the bright fringe is at position . For the second light with wavelength , the fringe separation changes, and the bright fringe is obtained at the same position on the screen. By equating the two equations for , we find: Simplifying, we get the value of as: Step 2: Final Answer.
Thus, the value of is .

Step 1: Condition for Coincidence of Bright Bands.
For the green and red light bands to coincide, their path difference should be equal. The fringe width for red and green light is proportional to the wavelength, and using this relationship, we can determine the value of for which the bright bands coincide. By equating the path differences and solving for , we find that .
Step 2: Final Answer.
Thus, the value of is 4.

Step 1: Formula for fringe width.
In Young’s double slit experiment, fringe width is given by where is the wavelength of light in the medium, is the distance between slit and screen, and is the separation between the slits.
Step 2: Effect of medium on wavelength.
When the experiment is performed in water instead of air, the wavelength of light decreases because where is the refractive index of water.
Step 3: Conclusion.
Since fringe width is directly proportional to wavelength, a decrease in wavelength leads to a decrease in fringe width.

Step 1: Fringe width formula in biprism experiment.
where is wavelength, is distance from slit to screen, and is separation between virtual sources.

Step 2: Writing expressions for both cases.

Step 3: Taking ratio.

Step 4: Conclusion.
The required ratio is .