367755
Four light waves are represented by (i) \({y_1} = {a_1}\sin (\omega t + \phi )\) (ii) \({y_2} = {a_2}\sin \omega t\) (iii) \({y_3} = {a_3}\sin 2(\omega t + \phi )\) (iv) \({y_4} = {a_4}\sin 2{\rm{ }}\omega t\) Interference fringes may be observed due to superposition of
1 (i) and (iii)
2 (i) and (ii)
3 (ii) and (iv)
4 (ii) and (iii)
Explanation:
The waves (i) & (ii) are of same frequencies and hence they are coherent. So interference fringes can be observed for these waves.
PHXII10:WAVE OPTICS
367756
The phenomenon of interference is shown by
1 Electromagnetic waves only
2 Longitudinal mechanical waves only
3 Transverse mechanical waves only
4 All the above types of waves
Explanation:
Interference is shown by electromagnetic as well as mechanical waves.
PHXII10:WAVE OPTICS
367757
To demonstrate the phenomenon of interference, we require two sources which emit radiations
1 of the same frequency and having definite phase relationship
2 of nearly the same frequency
3 of the same frequency
4 of different wavelength
Explanation:
For interference to occur two sources of light must be coherent (i.e. with same frequency and having a constant phase difference.) So correct option is (1)
PHXII10:WAVE OPTICS
367758
Two interfering light waves are mathematically represented as \(y_{1}=4 \sin \omega t\) and \(y_{2}=3 \cos (\omega t)\). The amplitude of the resultant wave on superposition is \(\left(y_{1}\right.\) and \(y_{2}\) are in \(C G S\) system)
1 \(5\;cm\)
2 \(7\;cm\)
3 \(1\;cm\)
4 Zero
Explanation:
\(A=\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \theta}\) Here \(\theta=90^{\circ}\) \(\Rightarrow A=\sqrt{a_{1}^{2}+a_{2}^{2}}=\sqrt{3^{2}+4^{2}}=5\)
367755
Four light waves are represented by (i) \({y_1} = {a_1}\sin (\omega t + \phi )\) (ii) \({y_2} = {a_2}\sin \omega t\) (iii) \({y_3} = {a_3}\sin 2(\omega t + \phi )\) (iv) \({y_4} = {a_4}\sin 2{\rm{ }}\omega t\) Interference fringes may be observed due to superposition of
1 (i) and (iii)
2 (i) and (ii)
3 (ii) and (iv)
4 (ii) and (iii)
Explanation:
The waves (i) & (ii) are of same frequencies and hence they are coherent. So interference fringes can be observed for these waves.
PHXII10:WAVE OPTICS
367756
The phenomenon of interference is shown by
1 Electromagnetic waves only
2 Longitudinal mechanical waves only
3 Transverse mechanical waves only
4 All the above types of waves
Explanation:
Interference is shown by electromagnetic as well as mechanical waves.
PHXII10:WAVE OPTICS
367757
To demonstrate the phenomenon of interference, we require two sources which emit radiations
1 of the same frequency and having definite phase relationship
2 of nearly the same frequency
3 of the same frequency
4 of different wavelength
Explanation:
For interference to occur two sources of light must be coherent (i.e. with same frequency and having a constant phase difference.) So correct option is (1)
PHXII10:WAVE OPTICS
367758
Two interfering light waves are mathematically represented as \(y_{1}=4 \sin \omega t\) and \(y_{2}=3 \cos (\omega t)\). The amplitude of the resultant wave on superposition is \(\left(y_{1}\right.\) and \(y_{2}\) are in \(C G S\) system)
1 \(5\;cm\)
2 \(7\;cm\)
3 \(1\;cm\)
4 Zero
Explanation:
\(A=\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \theta}\) Here \(\theta=90^{\circ}\) \(\Rightarrow A=\sqrt{a_{1}^{2}+a_{2}^{2}}=\sqrt{3^{2}+4^{2}}=5\)
367755
Four light waves are represented by (i) \({y_1} = {a_1}\sin (\omega t + \phi )\) (ii) \({y_2} = {a_2}\sin \omega t\) (iii) \({y_3} = {a_3}\sin 2(\omega t + \phi )\) (iv) \({y_4} = {a_4}\sin 2{\rm{ }}\omega t\) Interference fringes may be observed due to superposition of
1 (i) and (iii)
2 (i) and (ii)
3 (ii) and (iv)
4 (ii) and (iii)
Explanation:
The waves (i) & (ii) are of same frequencies and hence they are coherent. So interference fringes can be observed for these waves.
PHXII10:WAVE OPTICS
367756
The phenomenon of interference is shown by
1 Electromagnetic waves only
2 Longitudinal mechanical waves only
3 Transverse mechanical waves only
4 All the above types of waves
Explanation:
Interference is shown by electromagnetic as well as mechanical waves.
PHXII10:WAVE OPTICS
367757
To demonstrate the phenomenon of interference, we require two sources which emit radiations
1 of the same frequency and having definite phase relationship
2 of nearly the same frequency
3 of the same frequency
4 of different wavelength
Explanation:
For interference to occur two sources of light must be coherent (i.e. with same frequency and having a constant phase difference.) So correct option is (1)
PHXII10:WAVE OPTICS
367758
Two interfering light waves are mathematically represented as \(y_{1}=4 \sin \omega t\) and \(y_{2}=3 \cos (\omega t)\). The amplitude of the resultant wave on superposition is \(\left(y_{1}\right.\) and \(y_{2}\) are in \(C G S\) system)
1 \(5\;cm\)
2 \(7\;cm\)
3 \(1\;cm\)
4 Zero
Explanation:
\(A=\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \theta}\) Here \(\theta=90^{\circ}\) \(\Rightarrow A=\sqrt{a_{1}^{2}+a_{2}^{2}}=\sqrt{3^{2}+4^{2}}=5\)
367755
Four light waves are represented by (i) \({y_1} = {a_1}\sin (\omega t + \phi )\) (ii) \({y_2} = {a_2}\sin \omega t\) (iii) \({y_3} = {a_3}\sin 2(\omega t + \phi )\) (iv) \({y_4} = {a_4}\sin 2{\rm{ }}\omega t\) Interference fringes may be observed due to superposition of
1 (i) and (iii)
2 (i) and (ii)
3 (ii) and (iv)
4 (ii) and (iii)
Explanation:
The waves (i) & (ii) are of same frequencies and hence they are coherent. So interference fringes can be observed for these waves.
PHXII10:WAVE OPTICS
367756
The phenomenon of interference is shown by
1 Electromagnetic waves only
2 Longitudinal mechanical waves only
3 Transverse mechanical waves only
4 All the above types of waves
Explanation:
Interference is shown by electromagnetic as well as mechanical waves.
PHXII10:WAVE OPTICS
367757
To demonstrate the phenomenon of interference, we require two sources which emit radiations
1 of the same frequency and having definite phase relationship
2 of nearly the same frequency
3 of the same frequency
4 of different wavelength
Explanation:
For interference to occur two sources of light must be coherent (i.e. with same frequency and having a constant phase difference.) So correct option is (1)
PHXII10:WAVE OPTICS
367758
Two interfering light waves are mathematically represented as \(y_{1}=4 \sin \omega t\) and \(y_{2}=3 \cos (\omega t)\). The amplitude of the resultant wave on superposition is \(\left(y_{1}\right.\) and \(y_{2}\) are in \(C G S\) system)
1 \(5\;cm\)
2 \(7\;cm\)
3 \(1\;cm\)
4 Zero
Explanation:
\(A=\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \theta}\) Here \(\theta=90^{\circ}\) \(\Rightarrow A=\sqrt{a_{1}^{2}+a_{2}^{2}}=\sqrt{3^{2}+4^{2}}=5\)
