
In the Fraunhofer diffraction experiment, $ L $ is the distance between the screen and the obstacle, $ b $ is the size of the obstacle, and $ \lambda $ is the wavelength of incident light. The general condition for the applicability of Fraunhofer diffraction is
(A) $ \dfrac{{{b^2}}}{{L\lambda }} \gg 1 $
(B) $ \dfrac{{{b^2}}}{{L\lambda }} = 1 $
(C) $ \dfrac{{{b^2}}}{{L\lambda }} \ll 1 $
(D) $ \dfrac{{{b^2}}}{{L\lambda }} \ne 1 $
Answer
475.5k+ views
Hint :Fraunhofer diffraction experiment can only be used when the diffraction pattern is viewed at a long distance from the diffracting object. Hence, the value of $ L $ which is the distance between the screen and obstacle is very large.
Complete Step By Step Answer:
Diffraction can be defined as the bending or turning of the waves when it encounters an obstacle or passes through a thin slit, into the region of the shadow geometrically of the obstacle.
Diffraction can also be defined in simple words as the spreading of the waves at an opening or a slit.
Diffraction is of two types: Fresnel Diffraction and Fraunhofer Diffraction.
Fresnel Diffraction is used when the distance of the source of light and the display screen is within a finite distance of the obstacle.
As the waves emitted by a point source are spherical, the waves reaching the obstacle are spherical.
However, Fraunhofer diffraction is used when the light source and the screen are located at an infinite distance from the obstacle.
As the waves have traveled infinite distances, the radius of the spherical waves is infinite. For a section of the waves that are to be considered for the experiment, the waves can be considered as parallel waves, as a small part of the infinitely large circle is a straight line.
Hence, Fraunhofer diffraction is used when the distance of the obstacle from the source is very large, and also when it is viewed at the focal plane of the imaging lens.
Now, we are given that the distance between the screen and the obstacle is $ L $
As the diffraction occurring here is of Fraunhofer type, we can conclude that the value of $ L $ is near to infinite.
As it is present in the denominator, the overall value is very less than $ 1 $ .
Hence, to conduct the Fraunhofer diffraction experiment, the given value must be very less than $ 1 $ .
Hence, the correct answer is Option $ (C) $ .
Note :
Here, as it is given that the diffraction experiment conducted here is of the Fraunhofer type, we can derive the general condition required for the experiment to be true. If we use the Fresnel diffraction experiment, we also need the relation between the distance of source and obstacle and the size of the obstacle.
Complete Step By Step Answer:
Diffraction can be defined as the bending or turning of the waves when it encounters an obstacle or passes through a thin slit, into the region of the shadow geometrically of the obstacle.
Diffraction can also be defined in simple words as the spreading of the waves at an opening or a slit.
Diffraction is of two types: Fresnel Diffraction and Fraunhofer Diffraction.
Fresnel Diffraction is used when the distance of the source of light and the display screen is within a finite distance of the obstacle.
As the waves emitted by a point source are spherical, the waves reaching the obstacle are spherical.
However, Fraunhofer diffraction is used when the light source and the screen are located at an infinite distance from the obstacle.
As the waves have traveled infinite distances, the radius of the spherical waves is infinite. For a section of the waves that are to be considered for the experiment, the waves can be considered as parallel waves, as a small part of the infinitely large circle is a straight line.
Hence, Fraunhofer diffraction is used when the distance of the obstacle from the source is very large, and also when it is viewed at the focal plane of the imaging lens.
Now, we are given that the distance between the screen and the obstacle is $ L $
As the diffraction occurring here is of Fraunhofer type, we can conclude that the value of $ L $ is near to infinite.
As it is present in the denominator, the overall value is very less than $ 1 $ .
Hence, to conduct the Fraunhofer diffraction experiment, the given value must be very less than $ 1 $ .
Hence, the correct answer is Option $ (C) $ .
Note :
Here, as it is given that the diffraction experiment conducted here is of the Fraunhofer type, we can derive the general condition required for the experiment to be true. If we use the Fresnel diffraction experiment, we also need the relation between the distance of source and obstacle and the size of the obstacle.
Recently Updated Pages
Physics and Measurement Mock Test 2025 – Practice Questions & Answers

NCERT Solutions For Class 5 English Marigold - The Little Bully

NCERT Solutions For Class 12 Maths Three Dimensional Geometry Exercise 11.1

NCERT Solutions For Class 11 English Woven Words (Poem) - Ajamil And The Tigers

NCERT Solutions For Class 6 Hindi Durva - Bhaaloo

NCERT Solutions For Class 12 Physics In Hindi - Wave Optics

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE
