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1

AIEEE 2005

MCQ (Single Correct Answer)
When an unpolarized light of intensity $${{I_0}}$$ is incident on a polarizing sheet, the intensity of the light which does not get transmitted is
A
$${1 \over 4}\,{I_0}$$
B
$${1 \over 2}\,{I_0}$$
C
$${I_0}$$
D
zero

Explanation

$$I = {I_0}{\cos ^2}\theta $$

Intensity of polarized light $$ = {{{I_0}} \over 2}$$

$$ \Rightarrow $$ Intensity of untransmitted light $$ = {I_0} - {{{I_0}} \over 2} = {{{I_0}} \over 2}$$
2

AIEEE 2005

MCQ (Single Correct Answer)
If $${I_0}$$ is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled?
A
$$4{I_0}$$
B
$$2{I_0}$$
C
$${{{I_0}} \over 2}$$
D
$${I_0}$$

Explanation

$$I = {I_0}{\left( {{{\sin \phi } \over \phi }} \right)^2}\,\,$$ and $$\phi = {\pi \over \lambda }\left( {b\,\sin \theta } \right)$$

When the slit width is doubled, the amplitude of the wave at the center of the screen is doubled, so the intensity at the center is increased by a factor $$4.$$
3

AIEEE 2005

MCQ (Single Correct Answer)
A thin glass (refractive index $$1.5$$) lens has optical power of $$-5$$ $$D$$ in air. Its optical power in a liquid medium with refractive index $$1.6$$ will be
A
$$-1$$ $$D$$
B
$$1$$ $$D$$
C
$$-25$$ $$D$$
D
$$25$$ $$D$$

Explanation

$${1 \over {{f_a}}} = \left( {{{1.5} \over 1} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$${1 \over {{f_m}}} = \left( {{{{\mu _g}} \over {{\mu _m}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$

$${1 \over {{f_m}}} = \left( {{{1.5} \over {1.6}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Dividing $$(i)$$ by $$(ii)$$, $${{{f_m}} \over {{f_a}}} = \left( {{{1.5 - 1} \over {{{1.5} \over {1.6}} - 1}}} \right) = - 8$$

$${P_a} = - 5 = {1 \over {{f_a}}} \Rightarrow {f_a} = - {1 \over 5}$$

$$ \Rightarrow {f_m} = - 8 \times {f_a} = - 8 \times - {1 \over 5} = {8 \over 5}$$

$${P_m} = {\mu \over {{f_m}}} = {{1.6} \over 8} \times 5 = 1D$$
4

AIEEE 2005

MCQ (Single Correct Answer)
A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is
A
circle
B
hyperbola
C
parabola
D
straight line

Explanation

The shape of interference fringes formed on a screen in case of a monochromatic source is a straight line. Remember for double hole experiment a hyperbola is generated.

Questions Asked from Ray & Wave Optics

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