1
NEET 2016 Phase 2
MCQ (Single Correct Answer)
+4
-1
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio $${{{I_{max}} - {I_{\min }}} \over {{I_{max}} + {I_{min}}}}$$ will be
A
$${{\sqrt n } \over {n + 1}}$$
B
$${{2\sqrt n } \over {n + 1}}$$
C
$${{\sqrt n } \over {{{\left( {n + 1} \right)}^2}}}$$
D
$${{2\sqrt n } \over {{{\left( {n + 1} \right)}^2}}}$$
2
NEET 2016 Phase 2
MCQ (Single Correct Answer)
+4
-1
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aparture is illuminated normally by a parallel beam of wavelength 5 $$ \times $$ 10$$-$$5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is
A
0.10 cm
B
0.25 cm
C
0.20 cm
D
0.15 cm
3
NEET 2016 Phase 1
MCQ (Single Correct Answer)
+4
-1
In a diffraction pattern due to a single slit of width $$a$$, the first minimum is observed at an angle 30o when light of wavelength 5000 $$\mathop A\limits^ \circ $$ is incident on the slit. The first secondary maximum is observed at an angle of
A
sin$$-$$1$$\left( {{1 \over 2}} \right)$$
B
$${\sin ^{ - 1}}\left( {{3 \over 4}} \right)$$
C
$${\sin ^{ - 1}}\left( {{1 \over 4}} \right)$$
D
$${\sin ^{ - 1}}\left( {{2 \over 3}} \right)$$
4
NEET 2016 Phase 1
MCQ (Single Correct Answer)
+4
-1
The intensity at the maximum in a Young's double slit experiment is $$I$$0. Distance between two slits is d = 5$$\lambda $$, where $$\lambda $$ is the wavelength of light used in the expreriment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ?
A
$${3 \over 4}{I_0}$$
B
$${{{I_0}} \over 2}$$
C
I0
D
$${{{I_0}} \over 4}$$
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