Assume that the pupil of the human eye has a diameter of 4 . 00 mm and receives the light of wavelengths 5 . 0 × 10 - 7 m . b. Estimate the least distance between features on the Moon (a distance of 3 . 8 × 10 8 m away) that can be resolved by the human eye.

Given, The diameter of the pupil, D = 4 . 00 mm The wavelength of the light, λ = 5 . 0 × 10 - 7 m Distance, d = 3 . 8 × 10 8 m The least distance can be calculated by the given formula, θ = x d = 1 . 22 λ D x = 1 . 22 λ d D x = 1 . 22 × 5 . 0 × 10 - 7 4 . 00 × 10 - 3 × 3 . 8 × 10 8 x = 57 . 7 km Hence, the least distance is 57 . 7 km .

In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of n λ a . Justify this by suitably dividing the slit to bring out the cancellation.

Assume that at the angle of θ = n λ a , the intensity is zero. Consider that a single slit of width a is divided into n smaller slits. ∴ Width of each slit, a ' = a n The angle made by the each of these small slits can be written as, θ = n λ a = n λ a ' n = λ a ' If our initial assumption is correct, then at the angle of λ a ' the intensity produced by each of the smaller slits should be zero. This means, the total intensity due to all these slits will be zero. This result is in accordance with the initial assumption, so the assumption is justified.

E M B I B E

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 1

HARD

12th Karnataka Board

IMPORTANT

Earn 100

What is meant by limit of resolution and the resolving power of microscope?

Important Questions on Wave Optics

HARD

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 2

The semi vertical angle of cone of the rays incident on the objective of microscope is

20 ° .

If the wavelength of incident light ray is

6600

\overset{°}{A}

. Calculate the smallest distance between two points which can be just resolved.

MEDIUM

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 3

What is the minimum angular separation between two stars if telescope is used to observe them with an objective of circular aperture

20

cm ?

The wavelength of light used is

5900 \overset{°}{A}

.

HARD

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 4

What are resolution and resolving power?

HARD

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 5

How do you determine the resolving power of your eye?

MEDIUM

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 6

Assume that the pupil of the human eye has a diameter of $4.00 mm$ and receives the light of wavelengths $5.0 \times 10^{- 7} m$ .

b. Estimate the least distance between features on the Moon (a distance of $3.8 \times 10^{8} m$ away) that can be resolved by the human eye.

EASY

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 7

What is meant by fresnel distance

?

MEDIUM

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 8

In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of

n \frac{λ}{a}

. Justify this by suitably dividing the slit to bring out the cancellation.

MEDIUM

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Wave Optics>Exercise>Q 9

Two students are separated by a $7 m$ partition wall in a room $10 m$ high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.

Learn and Prepare for any exam you want !

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