Ray Optics and Optical Test

Result

Ray Optics and Optical Test - 44

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A cubical vessel has opaque walls. An observer (dark circle in figure below) is located such that she can see only the wall CD but not the bottom. Nearly to what height should water be poured so that she can see an object placed at the bottom at a distance of 10 cm from the corner C ? (Refractive index of water is 1.33.

A
10 cm

B
16 cm

C
27 cm

D
45 cm

Solution

$$\mu sin \, i = 1 \times sin 45^o$$
$$\dfrac{4}{3} sin \, i =\dfrac{1}{\sqrt{2}}$$
$$\dfrac{4}{3} \dfrac{y-10}{\sqrt{(y-10)^2+y^2}}=\dfrac{1}{\sqrt{2}}$$
$$y=27$$ (approximate calculation)
Question 2
1 / -0

A double convex lens of glass of refraction index $$\mu$$ is immersed in a medium of refractive index $$\mu_{1}$$. If a parallel beam emerges undeviated through the lens, then:

A
$$\mu = \mu_{1}$$

B
$$\mu = \dfrac {1}{\mu_{1}}$$

C
$$\mu > \mu_{1}$$

D
$$\mu < \mu_{1}$$

Solution

A parallel beam of light passes through the focus of a lens. For the light to remain undeviated, focus should lie at infinity or refraction must not occur so that light does not bend. For focus to be at infinity, radius should be infinitely large in which case a double convex lens cannot be formed. For refraction to not happen, refractive index of the medium of convex lens should be same as that of the medium. Hence, option A is correct.
Question 3
1 / -0

An isosceles glass prism with angles $${40}^{o}$$ is clamped over a tray of water in a position such that the base is just dipped in water. A ray of light incident normally on the inclined face suffers total internal reflection at the base. If the refractive index of water is $$1.33$$ then the condition imposed on the refractive index $$\mu$$ of the glass is :

A
$$\mu <2.07$$

B
$$\mu >2.07$$

C
$$\mu <1.74$$

D
$$\mu> 1.74$$

Solution

$$1.33\times \sin { { 90 }^{ o } } =\mu \sin { { 40 }^{ o } } $$
$$=\cfrac { 1.33\times 1 }{ \sin { { 40 }^{ o } } } =\mu =\cfrac { 1.33 }{ 3/5 } \approx \mu =\cfrac { 1.33\times 5 }{ 3 } \approx \mu $$
$$=\mu \approx 2.07\quad $$
$$\mu >2.07$$ (For TIR)
Question 4
1 / -0

A student sees the top edge and the bottom center C of a pool simultaneously from an angle $$\theta$$ above the horizontal as shown in the figure. The refractive index of water which fills up to the top edge of the pool is 4/3. If h/x = 7/4 then cos $$\theta$$ is

A
$$\dfrac{2}{7}$$

B
$$\dfrac{8}{3 \sqrt{45}}$$

C
$$\dfrac{8}{3 \sqrt{53}}$$

D
$$\dfrac{8}{21}$$

Solution

$$1\times\sin i=\mu\sin r$$
$$\sin(90-\theta)=\dfrac{4}{3}\sin r$$
$$\tan r=\dfrac{x}{2h}$$
$$\sin r=\dfrac{2}{\sqrt{53}}$$
$$\cos\theta=\dfrac{4}{3}\times\dfrac{2}{\sqrt{53}}=\dfrac{8}{3\sqrt{53}}$$
Question 5
1 / -0

A ray of light enters from a rarer to a denser medium. The angle of incidence is $$i$$. Then the reflected and refracted rays are mutually perpendicular to each other. The critical angle for the pair of media is

A
$$\sin ^{ -1 }{ \left( \tan { i } \right) } $$

B
$$\tan ^{ -1 }{ \left( \sin { i } \right) } $$

C
$$\sin ^{ -1 }{ \left( \cot { i } \right) } $$

D
$$\cos ^{ -1 }{ \left( \tan { i } \right) } $$

Solution

So $$i+r=\dfrac{\pi}{2}$$
$$sin{i}=\mu sinr$$
$$\mu=tan{i}$$
$$\mu sinc=1$$
$$c=sin^{-1}cot\,i$$
Question 6
1 / -0

A fish in water (refractive index $$n$$) looks at a bird vertically above in the air. If $$y$$ is the height of the bird and $$x$$ is the depth of the fish from the surface, then the distance of the bird as estimated by the fish is

A
$$x+y\left( 1-\dfrac { 1 }{ n } \right) $$

B
$$x+ny$$

C
$$x+y\left( 1+\dfrac { 1 }{ n } \right) $$

D
$$y+x\left( 1-\dfrac { 1 }{ n } \right) $$

Solution

Apparent height from surface =$$ny$$
total distance=$$x+ny$$
Question 7
1 / -0

A ray of white light of incident on a spherical water drop whose center is C as shown in figure. When observed from the opposite side, the emergent light.

A
Will be white and will emerge without deviating

B
Will be internally reflected

C
Will split into different colors such that the angles of deviation will be different for different colors

D
Will split into different colors such that the angles of deviation will be the same for all colors

Solution

Incident Ray is perpendicular ,So there won't be net deviation.
Question 8
1 / -0

The apparent flattening of the sun at sunset and sunrise is due to

A
refraction

B
diffraction

C
total internal reflection

D
interference

E
polarization

Solution

The apparent flattering of the sun at sunset and sunrise is due to refraction, because the sun becomes visible a little before actual sunrise and remains visible a little after actual sunset. because of atmosphere refraction of light.
Question 9
1 / -0

A light beam of diameter $$\sqrt{5}R$$ is incident symmetrically on a glass hemisphere of radius II and of refractive index $$n=\sqrt{3}$$. The radius of the beam at the base of hemisphere is:

A
$$\frac{R}{\sqrt{3}}$$

B
$$\frac{R}{2\sqrt{3}}$$

C
$$\frac{\sqrt{3}}{2} R$$

D
$$\frac{R}{2}$$

Solution

From the given ray diagram shown in figure below
$$\mu_1sin i=\mu_2 sin r$$
$$\Rightarrow 1 \times sin 60^o = \sqrt{3} \times sin \, r$$
$$\Rightarrow r=30^o$$
In $$\Delta MNP$$
$$tan 30^o =\frac{X}{(R/2)}$$
$$\Rightarrow X=\frac{R}{2\sqrt{3}}$$
From geometry of the figure
$$\Rightarrow r=\frac{\sqrt{3}}{2}R-X =\frac{\sqrt{3}}{2}R-\frac{R}{2\sqrt{3}}$$
$$=\left ( \frac{3-1}{2\sqrt{3}} \right )R=\frac{R}{\sqrt{3}}$$
Question 10
1 / -0

Light is incident normally on face AB of a prism as shown in figure. A liquid of refractive index $$\mu$$ is placed on the face AC of the prism. The prism is made of glass of refractive index 3/2. The limit of $$\mu$$ for which total internal reflection takes place on face AC is :

A
$$\mu <\dfrac{3\sqrt{3}}{4}$$

B
$$\mu > \sqrt{\dfrac{5}{2}}$$

C
$$\mu > \sqrt{2}$$

D
$$\mu < \sqrt{3}$$

Solution

Critical angle for glass-liquid :
$$_a\mu_g\, sin \, C=_a\mu _i\, sin 90^o$$
$$\Rightarrow \dfrac{3}{2} sin \, C=\mu \times 1$$
$$\Rightarrow sin \, C =\dfrac{2\mu}{3}$$
For Total Internal reflection on AC,
We must have $$60^o \geq C$$
$$\Rightarrow sin 60^o \geq sin \, C$$
$$\Rightarrow \dfrac{\sqrt{3}}{2}\geq \dfrac{2 \mu}{3}$$
Thus, we get $$\mu\leq \dfrac{3\sqrt{3}}{4}$$