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WAVE OPTICS

283499 A parallel beam of light of wavelength $λ$ is incident normally on a narrow slit. A diffraction pattern formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is

1

2 π

2

3 π

3

4 π

4

π λ

Explanation:

: Condition for diffraction minimum is
$Δ x = n λ$
$ϕ = 2 n π$
$ϕ = Phase difference$
$n = 2$
$Δ x = 2 λ$
$ϕ = 2 \times 2 \times π$
$= 4 π$

COMEDK 2013

WAVE OPTICS

283501 Assertion: The clouds in sky generally appear to be whitish.
Reason: Diffraction due to clouds is efficient in equal measure at all wavelengths.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

: The clouds in the sky appear white because the size of cloud particles is not so small to permit diffraction. So all the wavelength get refractive and it appear white.
From Brewster's law,
$i_{p} = \tan^{- 1} n$

AIIMS-2005

WAVE OPTICS

283505 A Single slit Fraunhoffer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength $6500 \AA$ ?

1

4400 \AA

2

4100 \AA

3

4642.8 \AA

4

9100 \AA

Explanation:

: For maxima in diffraction
$a \sin θ = (\frac{2 n + 1}{2}) λ$
For $3^{rd}$ secondary maxima using the unknown wavelength,
$a \sin θ_{2} = \frac{7 λ_{2}}{2}$
$\sin θ_{2} = \frac{7 λ_{2}}{2 a}$
For $2^{nd}$ secondary maxima using red light
$a \sin θ_{1} = \frac{5 λ_{1}}{2}$
$\sin θ_{1} = \frac{5 λ_{1}}{2 a}$
Since, these position coincide with each other for unknown wavelength and red light.
$\sin θ_{1} = \sin θ_{2}$
$\frac{5 λ_{1}}{2 a} = \frac{7 λ_{2}}{2 a}$
$λ_{2} = \frac{5}{7} λ_{1} = \frac{5}{7} \times 6500$
$λ_{2} = 4642.8 \AA$

AIIMS-2016

WAVE OPTICS

283507 The Fraunhoffer 'diffraction' pattern of a single slit is formed in the focal plane of a lens of focal length $1 m$ . The width of slit is $0.3 mm$ . If third minimum is formed at a distance of $5 mm$ from central maximum, then wavelength of light will

1

5000 \AA

2

2500 \AA

3

7500 \AA

4

8500 \AA

Explanation:

: From Fraunhofer diffraction case Path difference
We know that,
$Δ = a \sin θ$
$Δ = n λ$
From equation (i) and equation (ii), we get a $\sin θ = n λ$
$\frac{ay}{D} = 3 λ (∵ \sin θ = \frac{y}{D})$
$λ = \frac{ay}{3 D} = \frac{0.3 \times 10^{- 3} \times 5 \times 10^{- 3}}{3 \times 1} = 5 \times 10^{- 7} m$
$λ = 5000 \AA$

AIIMS-2015

WAVE OPTICS

283499 A parallel beam of light of wavelength $λ$ is incident normally on a narrow slit. A diffraction pattern formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is

1

2 π

2

3 π

3

4 π

4

π λ

Explanation:

: Condition for diffraction minimum is
$Δ x = n λ$
$ϕ = 2 n π$
$ϕ = Phase difference$
$n = 2$
$Δ x = 2 λ$
$ϕ = 2 \times 2 \times π$
$= 4 π$

COMEDK 2013

WAVE OPTICS

283501 Assertion: The clouds in sky generally appear to be whitish.
Reason: Diffraction due to clouds is efficient in equal measure at all wavelengths.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

: The clouds in the sky appear white because the size of cloud particles is not so small to permit diffraction. So all the wavelength get refractive and it appear white.
From Brewster's law,
$i_{p} = \tan^{- 1} n$

AIIMS-2005

WAVE OPTICS

283505 A Single slit Fraunhoffer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength $6500 \AA$ ?

1

4400 \AA

2

4100 \AA

3

4642.8 \AA

4

9100 \AA

Explanation:

: For maxima in diffraction
$a \sin θ = (\frac{2 n + 1}{2}) λ$
For $3^{rd}$ secondary maxima using the unknown wavelength,
$a \sin θ_{2} = \frac{7 λ_{2}}{2}$
$\sin θ_{2} = \frac{7 λ_{2}}{2 a}$
For $2^{nd}$ secondary maxima using red light
$a \sin θ_{1} = \frac{5 λ_{1}}{2}$
$\sin θ_{1} = \frac{5 λ_{1}}{2 a}$
Since, these position coincide with each other for unknown wavelength and red light.
$\sin θ_{1} = \sin θ_{2}$
$\frac{5 λ_{1}}{2 a} = \frac{7 λ_{2}}{2 a}$
$λ_{2} = \frac{5}{7} λ_{1} = \frac{5}{7} \times 6500$
$λ_{2} = 4642.8 \AA$

AIIMS-2016

WAVE OPTICS

283507 The Fraunhoffer 'diffraction' pattern of a single slit is formed in the focal plane of a lens of focal length $1 m$ . The width of slit is $0.3 mm$ . If third minimum is formed at a distance of $5 mm$ from central maximum, then wavelength of light will

1

5000 \AA

2

2500 \AA

3

7500 \AA

4

8500 \AA

Explanation:

: From Fraunhofer diffraction case Path difference
We know that,
$Δ = a \sin θ$
$Δ = n λ$
From equation (i) and equation (ii), we get a $\sin θ = n λ$
$\frac{ay}{D} = 3 λ (∵ \sin θ = \frac{y}{D})$
$λ = \frac{ay}{3 D} = \frac{0.3 \times 10^{- 3} \times 5 \times 10^{- 3}}{3 \times 1} = 5 \times 10^{- 7} m$
$λ = 5000 \AA$

AIIMS-2015

WAVE OPTICS

283499 A parallel beam of light of wavelength $λ$ is incident normally on a narrow slit. A diffraction pattern formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is

1

2 π

2

3 π

3

4 π

4

π λ

Explanation:

: Condition for diffraction minimum is
$Δ x = n λ$
$ϕ = 2 n π$
$ϕ = Phase difference$
$n = 2$
$Δ x = 2 λ$
$ϕ = 2 \times 2 \times π$
$= 4 π$

COMEDK 2013

WAVE OPTICS

283501 Assertion: The clouds in sky generally appear to be whitish.
Reason: Diffraction due to clouds is efficient in equal measure at all wavelengths.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

: The clouds in the sky appear white because the size of cloud particles is not so small to permit diffraction. So all the wavelength get refractive and it appear white.
From Brewster's law,
$i_{p} = \tan^{- 1} n$

AIIMS-2005

WAVE OPTICS

283505 A Single slit Fraunhoffer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength $6500 \AA$ ?

1

4400 \AA

2

4100 \AA

3

4642.8 \AA

4

9100 \AA

Explanation:

: For maxima in diffraction
$a \sin θ = (\frac{2 n + 1}{2}) λ$
For $3^{rd}$ secondary maxima using the unknown wavelength,
$a \sin θ_{2} = \frac{7 λ_{2}}{2}$
$\sin θ_{2} = \frac{7 λ_{2}}{2 a}$
For $2^{nd}$ secondary maxima using red light
$a \sin θ_{1} = \frac{5 λ_{1}}{2}$
$\sin θ_{1} = \frac{5 λ_{1}}{2 a}$
Since, these position coincide with each other for unknown wavelength and red light.
$\sin θ_{1} = \sin θ_{2}$
$\frac{5 λ_{1}}{2 a} = \frac{7 λ_{2}}{2 a}$
$λ_{2} = \frac{5}{7} λ_{1} = \frac{5}{7} \times 6500$
$λ_{2} = 4642.8 \AA$

AIIMS-2016

WAVE OPTICS

283507 The Fraunhoffer 'diffraction' pattern of a single slit is formed in the focal plane of a lens of focal length $1 m$ . The width of slit is $0.3 mm$ . If third minimum is formed at a distance of $5 mm$ from central maximum, then wavelength of light will

1

5000 \AA

2

2500 \AA

3

7500 \AA

4

8500 \AA

Explanation:

: From Fraunhofer diffraction case Path difference
We know that,
$Δ = a \sin θ$
$Δ = n λ$
From equation (i) and equation (ii), we get a $\sin θ = n λ$
$\frac{ay}{D} = 3 λ (∵ \sin θ = \frac{y}{D})$
$λ = \frac{ay}{3 D} = \frac{0.3 \times 10^{- 3} \times 5 \times 10^{- 3}}{3 \times 1} = 5 \times 10^{- 7} m$
$λ = 5000 \AA$

AIIMS-2015

WAVE OPTICS

283499 A parallel beam of light of wavelength $λ$ is incident normally on a narrow slit. A diffraction pattern formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is

1

2 π

2

3 π

3

4 π

4

π λ

Explanation:

: Condition for diffraction minimum is
$Δ x = n λ$
$ϕ = 2 n π$
$ϕ = Phase difference$
$n = 2$
$Δ x = 2 λ$
$ϕ = 2 \times 2 \times π$
$= 4 π$

COMEDK 2013

WAVE OPTICS

283501 Assertion: The clouds in sky generally appear to be whitish.
Reason: Diffraction due to clouds is efficient in equal measure at all wavelengths.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

: The clouds in the sky appear white because the size of cloud particles is not so small to permit diffraction. So all the wavelength get refractive and it appear white.
From Brewster's law,
$i_{p} = \tan^{- 1} n$

AIIMS-2005

WAVE OPTICS

283505 A Single slit Fraunhoffer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength $6500 \AA$ ?

1

4400 \AA

2

4100 \AA

3

4642.8 \AA

4

9100 \AA

Explanation:

: For maxima in diffraction
$a \sin θ = (\frac{2 n + 1}{2}) λ$
For $3^{rd}$ secondary maxima using the unknown wavelength,
$a \sin θ_{2} = \frac{7 λ_{2}}{2}$
$\sin θ_{2} = \frac{7 λ_{2}}{2 a}$
For $2^{nd}$ secondary maxima using red light
$a \sin θ_{1} = \frac{5 λ_{1}}{2}$
$\sin θ_{1} = \frac{5 λ_{1}}{2 a}$
Since, these position coincide with each other for unknown wavelength and red light.
$\sin θ_{1} = \sin θ_{2}$
$\frac{5 λ_{1}}{2 a} = \frac{7 λ_{2}}{2 a}$
$λ_{2} = \frac{5}{7} λ_{1} = \frac{5}{7} \times 6500$
$λ_{2} = 4642.8 \AA$

AIIMS-2016

WAVE OPTICS

283507 The Fraunhoffer 'diffraction' pattern of a single slit is formed in the focal plane of a lens of focal length $1 m$ . The width of slit is $0.3 mm$ . If third minimum is formed at a distance of $5 mm$ from central maximum, then wavelength of light will

1

5000 \AA

2

2500 \AA

3

7500 \AA

4

8500 \AA

Explanation:

: From Fraunhofer diffraction case Path difference
We know that,
$Δ = a \sin θ$
$Δ = n λ$
From equation (i) and equation (ii), we get a $\sin θ = n λ$
$\frac{ay}{D} = 3 λ (∵ \sin θ = \frac{y}{D})$
$λ = \frac{ay}{3 D} = \frac{0.3 \times 10^{- 3} \times 5 \times 10^{- 3}}{3 \times 1} = 5 \times 10^{- 7} m$
$λ = 5000 \AA$

AIIMS-2015

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