The intensity of diffracted light at any point on the screen is given by the formula I = I o sin α α 2 , using this formula give the conditions for central maximum, minima, and maxima. Where I ∘ = maximum intensity and α = πd sinθ λ .

For central maximum, the angle θ = 0 o which implies α = 0 o . Hence I = I o . For minima, α = ( 2 n ) π 2 which on simplifying, we get d sin θ = n λ For maxima, α = ( 2 n + 1 ) π 2 which on simplifying, we get d sin θ = ( 2 n + 1 ) λ / 2 where n is a natural number.

The intensity of diffracted light at any point on the screen is given by the formula I = I o sin α α 2 , using this formula give the conditions for central maximum, minima, and maxima. Where I ∘ = maximum intensity and α = πd sinθ λ .

For central maximum, the angle θ = 0 o which implies α = 0 o . Hence I = I o . For minima, α = ( 2 n ) π 2 which on simplifying, we get d sin θ = n λ For maxima, α = ( 2 n + 1 ) π 2 which on simplifying, we get d sin θ = ( 2 n + 1 ) λ / 2 where n is a natural number.

Draw a schematic diagram of Young's experiment and derive β = λ D d for the distance between two consecutive bright interference fringes.

Slits S 1 and S 2 are the coherent sources derived from the single slit S . A screen GG ' is placed at a distance D from the plane of the slits. The arrangement for Young's double-slit experiment is shown below: Let there be a point P at which a bright fringe occurs. Hence, the condition for maxima must be satisfied there and we can write the path difference: S 2 P - S 1 P = n λ ; n = 0 , 1 , 2 . . . Now: S 2 P 2 - S 1 P 2 = D 2 + x + d 2 2 - D 2 + x - d 2 2 = 2 x d where S 1 S 2 = d and OP = x . Thus: S 2 P - S 1 P = 2 x d S 2 P + S 1 P If x , d < < D then S 2 P + S 1 P can be replaced by 2 D . Hence: S 2 P - S 1 P = x d D So we will have constructive interference resulting in bright fringe when: x n d D = n λ ⇒ x n = n D λ d Hence, the distance between two bright fringes: β = x n + 1 - x n = D λ d This is the required expression for the two consecutive bright interference fringes.

State and prove Brewster's law.

According to brewster's law The tangent of the polarising angle is equal to the refractive index of the refracting medium at which partial reflection takes place that is tan i p = n . (i) When rays of light are allowed to fall on a transparent medium, then a part of the light gets reflected and the remaining gets refracted this phenomenon is called partial reflection. (ii) During partial reflection, the reflected light is partially polarised. At a certain angle of incidence, the reflected ray is completely polarised which is called polarised angle. (iii) Let AB be the incident light, let BD be the partially polarised refracted ray r. Let BC be the plane polarised reflected ray then, ∠ ABN = i , ∠ N ' BD = r , ∠ NBC = reflected angle According to the laws of reflection, the angle of incidence is always equal to the angle of reflection, therefore, ∠ i = ∠ i p ∠ ABN = ∠ i p = ∠ NBC And, ∠ DBC = 90 ° , now, from the given figure ∠ NBC + ∠ DBC + ∠ DBN ' = 180 ° = i p + 90 ° + r p = 180 ° i p = 90 ° - r p and r p = 90 ° - i p Now, according to the law of refraction or Snell's law sin i p sin r p = n = sin i p sin 90 ° - i p = sin i p cos i p = tan i p Or, n = tan i p , which is Brewster's law.

With proper diagram, explain the resolving power of a microscope and obtain its formula.

To obtain a good image, point sources must be resolved i.e., the point sources must be imaged such that their images are sufficiently far apart, such that the central maximum of the first image lies on the first minimum of the second( in its diffraction pattern) and vice versa. The resolving power is the inverse of the distance or angular separation between two objects which can be just resolved when viewed through the optical instrument. The value of the minimum distance between two objects, d varies directly as the wavelength λ and inversely as the cone angle of the light rays from anyone object entering the microscopei.e d ∝ λ 2 α where 2 α = cone angle. or the limit of resolution can be written as, d = λ 2 sinα . If the medium between the object and objective is a medium of refractive index μ , instead of air then we get, d = λ 2 μ sinα . Therefore, the resolving power of the microscope is given by, R.P of the microscope= 1 d = 2 μsinα λ . The term μ sinα is called the numerical aperture of the microscope.

A plane polarised light is incident normally on a tourmaline plate. Its E → vector makes an angle of 60 ° with the optic axis of the plate. Find the % difference between the initial and final maximum values of the E → vector.

According to the law of Malus, when a beam of completely plane polarised light is incident on an analyser, the resultant intensity of light transmitted from an analyser varies directly as the square of cosine angle between the plane of analyser and polariser. Let the initial intensity of the incident light is I 0 . And final intensity is of the light be I , then according to the Malu's law, I = I 0 cos 2 60 = I 0 1 2 2 = I 4 I 0 . Now, the difference between initial and final intensities, = I 0 - ( I 0 / 4 ) I 0 × 100 = 3 I 0 4 × 1 I 0 × 100 = 3 4 × 100 = 75 % . Thus the % difference is 75 % .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 10 - Wave Optics>Gujarat Board-2019>Q 9

EASY

12th Gujarat Board

IMPORTANT

Earn 100

The intensity of diffracted light at any point on the screen is given by the formula $I = I_{o} {(\frac{\sin α}{α})}^{2}$ , using this formula give the conditions for central maximum, minima, and maxima. Where $I_{\circ} =$ maximum intensity and $α = \frac{πd sinθ}{λ}$ .

Important Questions on Wave Optics

MEDIUM

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 10 - Wave Optics>Gujarat Board-2019>Q 10

In young's experiment, the distance between two slits and that between plates and the screen are

0.05 cm

and

1 m

, respectively. Find the distance between third bright and fifth dark fringes. Take the wavelength of the light equal to

5000 \overset{\circ}{A}

.

MEDIUM

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 10 - Wave Optics>Gujarat Board-2020>Q 13

Light of wavelength $6000 \overset{\circ}{A}$ is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of $100 inch$ ?

MEDIUM

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 10 - Wave Optics>Gujarat Board-2020>Q 14

Unpolarised light is incident on a plane glass surface. What should be the angle of incidence so that the reflected and refracted rays are perpendicular to each other?

MEDIUM

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 10 - Wave Optics>Gujarat Board-2020>Q 15

Draw a schematic diagram of Young's experiment and derive $β = \frac{λ D}{d}$ for the distance between two consecutive bright interference fringes.

HARD

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 10 - Wave Optics>Gujarat Board-2020>Q 16

The distance between the slits in Young's experiment is $0.1 mm$ . The perpendicular distance between the slits and screen is $1.5 m$ . The wavelength of the incident light is $6000 \overset{\circ}{A}$ . Calculate the distance between the third bright and fifth dark fringes, obtained on the screen.