Practice Start Test

Geometrical Optics

NEET Physics · 99 questions · Page 8 of 10 · Click an option or "Show Solution" to reveal answer

Q71
A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will
A become zero
B become infinite
C become small, but non-zero
D remain unchanged
Correct Answer
Option B
Solution
1f=(μ−1)(1R1−1R2){1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

Where

μ=μconvex lensμliquid\mu = {{{\mu _{convex\,lens}}} \over {{\mu _{liquid}}}}

= 1 ∴\therefore

1f=(1−1)(1R1−1R2)=0{1 \over f} = \left( {1 - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) = 0

⇒\Rightarrow f = ∞\infty

Q72
For the given incident ray as shown in figure, the condition of total internal refraction of this ray the required refractive index of prism will be
A 3+12{{\sqrt 3 + 1} \over 2}
B 2+12{{\sqrt 2 + 1} \over 2}
C 32\sqrt {{3 \over 2}}
D 76\sqrt {{7 \over 6}}
Correct Answer
Option C
Solution

Applying Snell’s law of refraction at A,

μ=sin⁡45∘sin⁡r\mu = {{\sin 45^\circ } \over {\sin r}}

⇒\Rightarrow r =

sin⁡−1(12μ){\sin ^{ - 1}}\left( {{1 \over {\sqrt 2 \mu }}} \right)

....(1) Applying the condition of total internal reflection at B, i c =

sin⁡−1(1μ){\sin ^{ - 1}}\left( {{1 \over \mu }} \right)

..........(2) where i c is the critical angle. From figure, r + i c =

π2{\pi \over 2}

⇒\Rightarrow

sin⁡−1(12μ)=π2−sin⁡−1(1μ){\sin ^{ - 1}}\left( {{1 \over {\sqrt 2 \mu }}} \right) = {\pi \over 2} - {\sin ^{ - 1}}\left( {{1 \over \mu }} \right)

⇒\Rightarrow

sin⁡−1(12μ)=cos⁡−1(1μ){\sin ^{ - 1}}\left( {{1 \over {\sqrt 2 \mu }}} \right) = {\cos ^{ - 1}}\left( {{1 \over \mu }} \right)

=

sin⁡−1(μ2−1μ){\sin ^{ - 1}}\left( {{{\sqrt {{\mu ^2} - 1} } \over \mu }} \right)

⇒\Rightarrow

12μ=μ2−1μ{{1 \over {\sqrt 2 \mu }} = {{\sqrt {{\mu ^2} - 1} } \over \mu }}

⇒\Rightarrow

μ2=1+12{\mu ^2} = 1 + {1 \over 2}

⇒\Rightarrow

μ=32\mu = \sqrt {{3 \over 2}}
Q73
Diameter of human eye lens is 2 mm. What will be the minimum distance between two points to resolve them, which are situated at a distance of 50 meter from eye. The wavelength of light is 5000 A∘\mathop A\limits^ \circ
A 2.32 m
B 4.28 mm
C 1.25 cm
D 12.48 cm
Correct Answer
Option C
Solution

For the aparture, limit of resolution :

yD≥λd{y \over D} \ge {\lambda \over d}

⇒\Rightarrow y ≥\ge

λDd{{\lambda D} \over d}

⇒\Rightarrow y

≥5×10−7×502×10−3\ge {{5 \times {{10}^{ - 7}} \times 50} \over {2 \times {{10}^{ - 3}}}}

≥\ge 1.25 cm

Q74
A bulb is located on a wall. Its image is to be obtained on a parallel wall with the help of convex lens. The lens is placed at a distance d ahead of second wall, then required focal length will be
A only d4{d \over 4}
B only d2{d \over 2}
C more than d4{d \over 4} but less than d2{d \over 2}
D less than d4{d \over 4}
Correct Answer
Option B
Solution

Using the lens formula

1f=1v−1u{1 \over f} = {1 \over v} - {1 \over u}

Given v = d, for equal size image |v| = |u| = d By sign convention u = -d ∴\therefore

1f=1d−1−d{1 \over f} = {1 \over d} - {1 \over { - d}}

⇒\Rightarrow f =

d2{d \over 2}
Q75
Optical fibre are based on
A total internal reflection
B less scattering
C refraction
D less absorption coefficient.
Correct Answer
Option A
Solution

Optical fibre are based on total internal reflection.

Q76
A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out is,
A ∞\infty
B 3 m
C 6 m
D 4 m
Correct Answer
Option B
Solution
θ=sin⁡−1(1μ)\theta = {\sin ^{ - 1}}\left( {{1 \over \mu }} \right)

=

sin⁡−1(35){\sin ^{ - 1}}\left( {{3 \over 5}} \right)

⇒\Rightarrow

sin⁡θ=35\sin \theta = {3 \over 5}

∴\therefore tanθ\theta =

34{3 \over 4}

=

r4{r \over 4}

⇒\Rightarrow r = 3 m

Q77
A bubble in glass slab (μ\mu = 1.5) when viewed from one side appears at 5 cm and 2 cm from other side, then thickness of slab is
A 3.75 cm
B 3 cm
C 10.5 cm
D 2.5 cm.
Correct Answer
Option C
Solution

From one side,

t−x5=1.5{{t - x} \over 5} = 1.5

From other side,

x2{x \over 2}

= 1.5 ⇒\Rightarrow x = 3 ∴\therefore

t−35=1.5{{t - 3} \over 5} = 1.5

⇒\Rightarrow t = 10.5 cm

Q78
A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirror will be
A 12 feet
B 3 feet
C 6 feet
D any length
Correct Answer
Option B
Solution

The minimum mirror length should be half of the height of man.

Q79
For a plano convex lens (μ\mu = 1.5) has radius of curvature 10 cm. It is silvered on its plane surface. Find focal length after silvering
A 10 cm
B 20 cm
C 15 cm
D 25 cm
Correct Answer
Option A
Solution
1f=(μ−1)(1R1−1R2){1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

=

(1.5−1)(1∞−1−10)\left( {1.5 - 1} \right)\left( {{1 \over \infty } - {1 \over { - 10}}} \right)

=

0.5(110)0.5\left( {{1 \over {10}}} \right)

⇒\Rightarrow f = 20 cm When plane surface is silvered, F =

f2=202{f \over 2} = {{20} \over 2}

= 10 cm

Q80
Rainbow is formed due to
A scattering and refraction
B internal reflection and dispersion
C reflection only
D diffraction and dispersion.
Correct Answer
Option B
Solution

Rainbow is formed due to combination of internal reflection and dispersion.

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