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Geometrical Optics

JEE Physics · 130 questions · Page 3 of 13 · Click an option or "Show Solution" to reveal answer

Q21
A person has been using spectacles of power −1.0-1.0 dioptre for distant vision and a separate reading glass of power 2.02.0 dioptres. What is the least distance of distinct vision for this person :
A 50 cm
B 40 cm
C 30 cm
D 10 cm
Correct Answer
Option A
Solution

u = 25 cm

f=12f=\frac{1}{2}

m = 50 cm

1v−1u=1f{1 \over v} - {1 \over u} = {1 \over f}
⇒1v+125=−150\Rightarrow {1 \over v} + {1 \over {25}} = - {1 \over {50}}
1v=−150{1 \over v} = - {1 \over {50}}
⇒u=−50\Rightarrow u = - 50

cm

Q22
There is a small source of light at some depth below the surface of water (refractive index = 43{4 \over 3}) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly) : [Use the fact that surface area of a spherical cap of height h and radius of curvature r is 2π\pi rh]:
A 17%
B 34%
C 50%
D 21%
Correct Answer
Option A
Solution
43{4 \over 3}

sin θ\theta = 1sin90o sin θ\theta =

34{3 \over 4}

cos θ\theta =

74{{\sqrt 7 } \over 4}

Surface area in solid angle d

Ω\Omega

= 2π\piR2(1 - cos θ\theta) = 2π\piR2(1 -

74{{\sqrt 7 } \over 4}

) Percentage of light =

2πR2(1−74)4πR2{{2\pi {R^2}\left( {1 - {{\sqrt 7 } \over 4}} \right)} \over {4\pi {R^2}}}

×\times 100% =

4−78{{{4 - \sqrt 7 } \over 8}}

×\times 100% = 17%

Q23
A short straight object of height 100 cm lies before the central axis of a spherical mirror whose focal length has absolute value | f | = 40 cm. The image of object produced by the mirror is of height 25 cm and has the same orientation of the object. One may conclude from the information :
A Image is real, same side of convex mirror.
B Image is virtual, opposite side of convex mirror.
C Image is virtual, opposite side of concave mirror.
D Image is real, same side of concave mirror.
Correct Answer
Option B
Solution

Same orientation so image is virtual.

It is combination of real object and virtual image using height, it is possible only from convex mirror.

Q24
An object is placed at the focus of concave lens having focal length f. What is the magnification and distance of the image from the optical centre of the lens?
A 1, ∞\infty
B Very high, ∞\infty
C 12{1 \over 2}, f2{f \over 2}
D 14{1 \over 4}, f4{f \over 4}
Correct Answer
Option C
Solution

U = −-f

1V−1U=1−f⇒1V=−2f{1 \over V} - {1 \over U} = {1 \over { - f}} \Rightarrow {1 \over V} = - {2 \over f}
V=−f2V = {{ - f} \over 2}
m=VU=12m = {V \over U} = {1 \over 2}

distance =

f2{f \over 2}

Option (c)

Q25
A glass tumbler having inner depth of 17.5 cm is kept on a table. A student starts pouring water (μ\mu = 4/3) into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?
A 11.7 cm
B 10 cm
C 7.5 cm
D 8.75 cm
Correct Answer
Option B
Solution

Consider the actual height of the tumbler be H.

The refractive index of the water, μ\mu = 4/3 The refractive index of the air, μ\mu = 1 As we know that,

μwater=HrealHapparent⇒43=HHapparent{\mu _{water}} = {{{H_{real}}} \over {{H_{apparent}}}} \Rightarrow {4 \over 3} = {H \over {{H_{apparent}}}}
⇒Happarent=3H4\Rightarrow {H_{apparent}} = {{3H} \over 4}

Height of air observed by observer = 17.5 −- H According to question, both height observed by observer is same.

3H4{{3H} \over 4}

= 17.5 −- H ⇒\Rightarrow H = 10 cm Option (b)

Q26
The image formed by an objective of a compound microscope is
A virtual and diminished
B real an diminished
C real and enlarged
D virtual and enlarged
Correct Answer
Option C
Solution

A real, inverted and enlarged image of the object is formed by the objective lens of a compound microscope.

Q27
To get three images of a single object, one should have two plane mirrors at an angle of
A 60∘{60^ \circ }
B 90∘{90^ \circ }
C 120∘{120^ \circ }
D 30∘{30^ \circ }
Correct Answer
Option B
Solution

When

θ=90∘\theta = {90^ \circ }

then

360θ=36090=4{{360} \over \theta } = {{360} \over {90}} = 4

is an even number. The number of images formed is given by

n=360θ−1=36090−1=4−1=3n = {{360} \over \theta } - 1 = {{360} \over {90}} - 1 = 4 - 1 = 3
Q28
Consider a light ray travelling in air is incident into a medium of refractive index 2n\sqrt{2n}. The incident angle is twice that of refracting angle. Then, the angle of incidence will be :
A sin⁡−1(n){\sin ^{ - 1}}\left( {\sqrt n } \right)
B cos⁡−1(n2){\cos ^{ - 1}}\left( {\sqrt {{n \over 2}} } \right)
C sin⁡−1(2n){\sin ^{ - 1}}\left( {\sqrt {2n} } \right)
D 2cos⁡−1(n2)2{\cos ^{ - 1}}\left( {\sqrt {{n \over 2}} } \right)
Correct Answer
Option D
Solution

According to the law,

1×sin⁡θ=2n×sin⁡(θ2)1 \times \sin \theta = \sqrt {2n} \times \sin \left( {{\theta \over 2}} \right)
⇒cos⁡θ2=n2\Rightarrow \cos {\theta \over 2} = \sqrt {{n \over 2}}
⇒θ=2cos⁡−1(n2)\Rightarrow \theta = 2{\cos ^{ - 1}}\left( {\sqrt {{n \over 2}} } \right)
Q29
An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be :
A 1.16 × \times 10–3 m/s towards the lens
B 2.26 × \times 10–3 m/s away from the lens
C 3.22 × 10–3 m/s towards the lens
D 0.92 × \times 10−-3 m/s away from the lens
Correct Answer
Option A
Solution

From lens equation

1v−1u=1f{1 \over v} - {1 \over u} = {1 \over f}
1v−1(−20)=1(.3)=103{1 \over v} - {1 \over {\left( { - 20} \right)}} = {1 \over {\left( {.3} \right)}} = {{10} \over 3}
1v=103−120{1 \over v} = {{10} \over 3} - {1 \over {20}}
1v=19760;v=60197{1 \over v} = {{197} \over {60}};v = {{60} \over {197}}

m =

(vu)\left( {{v \over u}} \right)

=

(60197)20{{\left( {{{60} \over {197}}} \right)} \over {20}}

velocity of image wrt. to lens is given by vI/L = m2vO/L direction of velocity of image is same as that of object vO/L = 5 m/s

Q30
A thin glass (refractive index 1.51.5) lens has optical power of −5-5 DD in air. Its optical power in a liquid medium with refractive index 1.61.6 will be
A −1-1 DD
B 11 DD
C −25-25 DD
D 2525 DD
Correct Answer
Option B
Solution
1fa=(1.51−1)(1R1−1R2)            ...(i){1 \over {{f_a}}} = \left( {{{1.5} \over 1} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)
1fm=(μgμm−1)(1R1−1R2){1 \over {{f_m}}} = \left( {{{{\mu _g}} \over {{\mu _m}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)
1fm=(1.51.6−1)(1R1−1R2)            ...(ii){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)(i)

by

(ii)(ii)

,

fmfa=(1.5−11.51.6−1)=−8{{{f_m}} \over {{f_a}}} = \left( {{{1.5 - 1} \over {{{1.5} \over {1.6}} - 1}}} \right) = - 8
Pa=−5=1fa⇒fa=−15{P_a} = - 5 = {1 \over {{f_a}}} \Rightarrow {f_a} = - {1 \over 5}
⇒fm=−8×fa=−8×−15=85\Rightarrow {f_m} = - 8 \times {f_a} = - 8 \times - {1 \over 5} = {8 \over 5}
Pm=μfm=1.68×5=1D{P_m} = {\mu \over {{f_m}}} = {{1.6} \over 8} \times 5 = 1D
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