Diffraction by a Circular Aperture MCQ Quiz - Objective Question with Answer for Diffraction by a Circular Aperture - Download Free PDF

CONCEPT:

Fresnel’s Half Period Zone (HPZ):

• According to Fresnel’s the entire wavefront can be divided into a large number of parts of zones which are known as Fresnel’s half-period zones (HPZ’s).
• The resultant effect at any point on the screen is due to the combined effect of all the secondary waves from the various zones.

EXPLANATION:

• The radius of half period zone is:

\(⇒ r_n=\sqrt{ndλ}\)

Where n = number of zones, d = distance from point P to the wavefront, and λ = wavelength

• The Area of the nth HPZ is given by 

An = πdλ 

Calculation:

Given d = 1m, λ = 7000 Å = 7 × 10^-7 m;

Now the are of the HPZ will be 

A = π × 1 × 7 × 10^-7 m²;

⇒ A = 21.99 × 10^-7 m²;

Concept:

• Resolving Power of Microscope: The resolving power of a microscope is the ability of the microscope to show separate, images of two-point objects lying close to each other.
  • \(RP = {2\mu Sin\theta \over λ }\); 
  • where λ = Wavelength of light used to illuminate the object, μ = Refractive index of the medium between object and objective, and μ sinθ = Numerical aperture.
• The numerical aperture is the measure of its ability to gather light and to resolve fine specimen detail while working at a fixed object (or specimen) distance.

Explanation:

The wavelength of red light = 620 to 750 nm

The wavelength of blue light = 450 to 495 nm

As we know the Resolving power of a microscope is given by;

\(RP = {2\mu Sin\theta \over λ }\)

⇒ \(RP ∝ {1 \over λ}\) 

Here if the red light is replaced by blue light, so we are decreasing the wavelength.

Therefore the resolving power of the microscope will increase.

CONCEPT:

Fresnel’s Half Period Zone (HPZ):

• According to Fresnel’s the entire wavefront can be divided into a large number of parts of zones which are known as Fresnel’s half-period zones (HPZ’s).
• The resultant effect at any point on the screen is due to the combined effect of all the secondary waves from the various zones.

EXPLANATION:

• The radius of half period zone is:

\(⇒ r_n=\sqrt{ndλ}\)

Where n = number of zones, d = distance from point P to the wavefront, and λ = wavelength

Calculation:

Given n = 4, d = 4 m, λ = 6400 A°;

Now the radius of the fourth half-period zone will be 

\(⇒ r_4=\sqrt{4×4×6400×10^{-10}}\)

⇒ r₄ = 32 × 10^-4 m;

CONCEPT:

Fresnel’s Half Period Zone (HPZ):

• According to Fresnel’s the entire wavefront can be divided into a large number of parts of zones which are known as Fresnel’s half-period zones (HPZ’s).
• The resultant effect at any point on the screen is due to the combined effect of all the secondary waves from the various zones.

EXPLANATION:

• The radius of half period zone is:

\(⇒ r_n=\sqrt{ndλ}\)

Where n = number of zones, d = distance from point P to the wavefront, and λ = wavelength

• The Area of the nth HPZ is given by 

An = πdλ 

Calculation:

Given d = 1m, λ = 7000 Å = 7 × 10^-7 m;

Now the are of the HPZ will be 

A = π × 1 × 7 × 10^-7 m²;

⇒ A = 21.99 × 10^-7 m²;

CONCEPT:

Fresnel’s Half Period Zone (HPZ):

• According to Fresnel’s the entire wavefront can be divided into a large number of parts of zones which are known as Fresnel’s half-period zones (HPZ’s).
• The resultant effect at any point on the screen is due to the combined effect of all the secondary waves from the various zones.

EXPLANATION:

• The radius of half period zone is:

\(⇒ r_n=\sqrt{ndλ}\)

Where n = number of zones, d = distance from point P to the wavefront, and λ = wavelength

Calculation:

Given n = 4, d = 4 m, λ = 6400 A°;

Now the radius of the fourth half-period zone will be 

\(⇒ r_4=\sqrt{4×4×6400×10^{-10}}\)

⇒ r₄ = 32 × 10^-4 m;

CONCEPT:

Fresnel’s Half Period Zone (HPZ):

• According to Fresnel’s the entire wavefront can be divided into a large number of parts of zones which are known as Fresnel’s half-period zones (HPZ’s).
• The resultant effect at any point on the screen is due to the combined effect of all the secondary waves from the various zones.

EXPLANATION:

• The radius of half period zone is:

\(⇒ r_n=\sqrt{ndλ}\)

Where n = number of zones, d = distance from point P to the wavefront, and λ = wavelength

• The Area of the nth HPZ is given by 

An = πdλ 

Calculation:

Given d = 1m, λ = 7000 Å = 7 × 10^-7 m;

Now the are of the HPZ will be 

A = π × 1 × 7 × 10^-7 m²;

⇒ A = 21.99 × 10^-7 m²;

CONCEPT:

Fresnel’s Half Period Zone (HPZ):

• According to Fresnel’s the entire wavefront can be divided into a large number of parts of zones which are known as Fresnel’s half-period zones (HPZ’s).
• The resultant effect at any point on the screen is due to the combined effect of all the secondary waves from the various zones.

EXPLANATION:

• The radius of half period zone is:

\(⇒ r_n=\sqrt{ndλ}\)

Where n = number of zones, d = distance from point P to the wavefront, and λ = wavelength

Calculation:

Given n = 4, d = 4 m, λ = 6400 A°;

Now the radius of the fourth half-period zone will be 

\(⇒ r_4=\sqrt{4×4×6400×10^{-10}}\)

⇒ r₄ = 32 × 10^-4 m;

CONCEPT:

Fresnel’s Half Period Zone (HPZ):

• According to Fresnel’s the entire wavefront can be divided into a large number of parts of zones which are known as Fresnel’s half-period zones (HPZ’s).
• The resultant effect at any point on the screen is due to the combined effect of all the secondary waves from the various zones.

EXPLANATION:

• The radius of half period zone is:

\(⇒ r_n=\sqrt{ndλ}\)

Where n = number of zones, d = distance from point P to the wavefront, and λ = wavelength

• The Area of the nth HPZ is given by 

An = πdλ 

Calculation:

Given d = 1m, λ = 7000 Å = 7 × 10^-7 m;

Now the are of the HPZ will be 

A = π × 1 × 7 × 10^-7 m²;

⇒ A = 21.99 × 10^-7 m²;

CONCEPT:

Fresnel’s Half Period Zone (HPZ):

• According to Fresnel’s the entire wavefront can be divided into a large number of parts of zones which are known as Fresnel’s half-period zones (HPZ’s).
• The resultant effect at any point on the screen is due to the combined effect of all the secondary waves from the various zones.

EXPLANATION:

• The radius of half period zone is:

\(⇒ r_n=\sqrt{ndλ}\)

Where n = number of zones, d = distance from point P to the wavefront, and λ = wavelength

• The Area of the nth HPZ is given by 

An = πdλ 

Calculation:

Given d = 1m, λ = 7000 Å = 7 × 10^-7 m;

Now the are of the HPZ will be 

A = π × 1 × 7 × 10^-7 m²;

⇒ A = 21.99 × 10^-7 m²;

Concept:

• Resolving Power of Microscope: The resolving power of a microscope is the ability of the microscope to show separate, images of two-point objects lying close to each other.
  • \(RP = {2\mu Sin\theta \over λ }\); 
  • where λ = Wavelength of light used to illuminate the object, μ = Refractive index of the medium between object and objective, and μ sinθ = Numerical aperture.
• The numerical aperture is the measure of its ability to gather light and to resolve fine specimen detail while working at a fixed object (or specimen) distance.

Explanation:

The wavelength of red light = 620 to 750 nm

The wavelength of blue light = 450 to 495 nm

As we know the Resolving power of a microscope is given by;

\(RP = {2\mu Sin\theta \over λ }\)

⇒ \(RP ∝ {1 \over λ}\) 

Here if the red light is replaced by blue light, so we are decreasing the wavelength.

Therefore the resolving power of the microscope will increase.