Ray Optics and Optical Test

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Ray Optics and Optical Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

A fish in water sees an object $$24\ cm$$ above the surface of water. The height of object above the surface of water as it appear to the fish is

A
$$32\ cm$$

B
$$8\ cm$$

C
$$24\ cm$$

D
$$18\ cm$$

Solution

$$\mu_{medium} = \dfrac{h_{medium}}{h}$$
$$\dfrac{4}{3} = \dfrac{h^1}{24} \\ $$
$$h^` = \dfrac{4}{3} \times 24 \\ $$
$$h^` = 32 cm $$
Hence option 'A' is correct.
Question 2
1 / -0

The image of an object formed by a device is always virtual and small. The device may be

A
A glass plate

B
A concave mirror

C
A convex lens

D
A concave lens

Solution

Concave lenses always produce images that are upright, virtual, reduced in size, and located on the object's side of the lens.
Therefore, the image of an object formed by a concave lens is always virtual and small.
Question 3
1 / -0

A real image of a distant object is formed by a plano-convex lens on its principal axis. Spherical aberration is :

A
absent.`

B
smaller, if the curved surfaces of the lens face an object.

C
smaller, if the plane surface of the lens faces the object.

D
same, whichever side of the lens faces the object.

Solution

$$\textbf{Explanation}$$
$$\bullet$$The spherical aberration for a plano-convex lens will be minimum when the curved surface faces the object and then the light from the object will fall on the curved surface.
$$\bullet$$This is because the total deviation of the ray is distributed between the two surfaces. Th total deviation is shared by both the surfaces.

Hence, the spherical aberration is smaller if the curved surface of the lens faces the object.
The correct answer is option (B).
Question 4
1 / -0

A plane mirror is placed along the x-axis facing negative y-axis. The mirror is fixed. A point object is moving with $$3\hat { i } +4\hat { j } $$ in front of the plane mirror. The relative velocity of image with respect to its object is

A
$$-8\hat { j } $$

B
$$8\hat { j } $$

C
$$3\hat { i } -4\hat { j } $$

D
$$-6\hat { i } $$

Solution

$$\overrightarrow { { V }_{ I } } =3\hat { i } +4\hat { j } \\ \overrightarrow { { V }_{ R } } =3\hat { i } -4\hat { j } \\ \therefore \overrightarrow { { V }_{ R } } -\overrightarrow { { V }_{ I } } =-8\hat { j } $$
Question 5
1 / -0

Two mirrors at an angle $$\theta $$ produce 5 images of a point. The number of images produced when $$\theta $$ is decreased to $$\theta - 30 $$ is

A
$$9$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

Number of images=$$\cfrac{360}{\theta}-1$$
$$\Longrightarrow \cfrac { 360 }{ \theta } -1=5\\ \therefore \theta =60°$$
If now, $$\theta =30°$$
then, the number of images=$$\cfrac{360}{\theta}-1=\cfrac{360}{30^{\circ}}=12-1=11$$
Question 6
1 / -0

A glass prism of refractive index $$1.5$$ is immersed in water (refractive index $$4/3$$). A light beam incident normally on the face:$$AB$$ as shown in the figure is totally reflected to reach the face $$BC$$ if:

A
$$\sin { \theta } >8/9$$

B
$$2/3<\sin { \theta } <8/9$$

C
$$\sin { \theta } \le 2/3$$

D
$$none\ of\ the\ above$$

Solution

$$ \begin{array}{l} \text { Condition for } \mathbb{Refraction} \quad \mu_{1} \sin \theta_{1}, \quad>\mu_{2} \sin \theta_{2} \\ 1.5 \sin \theta \cdot>\dfrac{4}{3} \sin (90) \\ \sin \theta>\dfrac{8}{9} \\ \text { So option } A \text { is correct } \end{array} $$
Question 7
1 / -0

A light ray is incident at an angle of $${45}^{o}$$ with the normal to a $$\sqrt {2}cm$$ thick plate ($$\mu =\sqrt2$$). Find the shift in the path of the light as it emerges out from the plate.

A
$$0.42cm$$

B
$$0.57cm$$

C
$$0.62cm$$

D
$$0.67cm$$

Solution

Lateral displacement is given by
$$x=\cfrac { t\sin { \left( i-r \right) } }{ \cos { r } } $$
Using Snell's law
$$1\times \sin { i=\mu \times \sin { r } } \\ \therefore \sin { r } =\cfrac { 1 }{ \sqrt { 2 } } \times \cfrac { 1 }{ \sqrt { 2 } } =\cfrac { 1 }{ 2 } \\ \therefore r=30°$$
Hence, $$x=\cfrac { \sqrt { 2 } \times \sin { \left( 45°-30° \right) } }{ \cos { 30° } } \\ \quad =\cfrac { \sqrt { 2 } \times \cfrac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } } }{ \cfrac { \sqrt { 3 } }{ 2 } } =0.42cm$$
Question 8
1 / -0

In the given situation, a spherical transparent surface of radius $$R$$ is placed. The refractive index of sphere is such that image for an object kept at distance $$R$$ is formed at the same distance of the opposite side. What is refractive index of sphere?

A
$$1.5$$

B
$$1.75$$

C
$$1.25$$

D
$$2$$

Solution

From the symmetry of the figure, ray inside the sphere is parallel to principal axis. Taking refraction at $$A$$.

$$\dfrac{\mu_2}{v}-\dfrac{\mu_1}{u}=\dfrac{\mu_2-\mu_1}{R}$$

$$\dfrac{\mu}{\infty}-\dfrac{1}{-R}=\dfrac{\mu-1}{R}$$

$$\dfrac 1R=\dfrac{\mu -1}{R}\implies \mu -1=1$$

$$\implies \mu=2$$
Question 9
1 / -0

An air bubble is inside water. The refractive index of water is $$4/3$$. At what distance from the air bubble should a point object be placed so as to form a real image at the same distance from the bubble?

A
$$2R$$

B
$$3R$$

C
$$4R$$

D
The air bubble cannot form a real image

Solution

A real image can not be formed in the given case.
Bubbles are formed, when air is trapped in water, because the air is less dense, the air bubble will behave as a diverging lens due to its bulging curvature and a diverging lens always forms a virtual and erect image.
Question 10
1 / -0

An equilateral prism deviates a ray through $$45^{\circ}$$ for the two angles of incidence differing by $$20^{\circ}.$$ The angle of incidence is

A
$$62.5^{\circ}$$

B
$$42.5^{\circ}$$

C
Both are correct

D
Both are wrong

Solution

First angle of incidence $$=i_1=i$$
Second angle of incidence $$i_2=i_1+20^\circ$$
Angle of deviation, $$d=i_1+i_2-A$$ where A is angle of prism.
$$45=2i+20-60$$
$$i=42.5$$
$$i_2=i+20=62.5$$
Hence both are correct.