Ray Optics and Optical Test

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Ray Optics and Optical Test - 45

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Question 1
1 / -0

There is a small block in the centre C of solid glass sphere of refractive index $$\mu$$. When seen from outside the block will appear to be located :

A
Away from C for all values of $$\mu$$

B
at C for all values of $$\mu$$

C
at C for $$\mu=1.5$$ but away from C for $$\mu \neq 1.5$$

D
at C only for$$ \sqrt{2} \leq\mu\leq 1.5$$

Solution

If the object is placed at the centre of the sphere, all the rays coming from the block will be normal to the refracting surface. So angle of incident $$0^o$$, there for ray pass straight without any deviation.
Question 2
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A ray of light is incident on a surface of glass slab at an angle $$45^o$$. If the lateral shift produced per unit thickness is $$1/\sqrt{3}$$, then the angle of refraction produced is?

A
$$\displaystyle \tan^{-1}\left(\displaystyle\frac{\sqrt{3}}{2}\right)$$

B
$$\displaystyle\tan^{-1}\left(\displaystyle 1-\sqrt{\frac{2}{3}}\right)$$

C
$$\displaystyle\sin^{-1}\left(\displaystyle 1-\sqrt{\frac{2}{3}}\right)$$

D
$$\displaystyle\tan^{-1}\left(\displaystyle\sqrt{\frac{2}{3-1}}\right)$$

Solution

Lateral shift, $$Y=\displaystyle\frac{t\sin(i-r)}{\cos r}$$

$$\displaystyle\frac{Y}{t}=\frac{\sin (i-r)}{\cos r}$$

$$=\displaystyle\frac{\sin i\cos r-\cos i\sin r}{\cos r}$$

[using identity, $$\sin(A-B)=\sin A\cdot \cos B-\cos A\cdot \sin B$$]

$$=\sin i-\cos i\tan r$$

$$\Rightarrow \displaystyle\frac{1}{\sqrt{3}}=\sin 45^o-\cos 45^o\tan r=\frac{1}{\sqrt{2}}(1-\tan r)$$

$$\Rightarrow \sqrt{\displaystyle \frac{2}{3}}=1-\tan r$$

$$\Rightarrow \tan r=1-\sqrt{\displaystyle\frac{2}{3}}$$

$$\therefore r=\tan^{-1}\left(1-\displaystyle\sqrt{\frac{2}{3}}\right)$$.
Question 3
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A narrow white light beam fails to converge at a point after going through a converging lens. This defect is known as :

A
Chromatic aberration

B
Diffraction

C
Polarisation

D
Spherical aberration

Solution

This defect is known as chromatic aberration. It occurs because the lenses have different refractive indices for different wave lengths of light. Chromatic abberation itself manifests as "fringes" of colour along boundaries that seperate dark and bright parts of the image, because each colour in the optical spectrum can not be focused at a single common point.
Question 4
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A straight rod is partially immersed in water $$(\mu = 1.33)$$. Its submerged portion appears to be inclined at $$45^o$$ with the surface when viewed vertically from air. The actual inclination of the rod is?

A
$$32.14^o$$

B
$$45^o$$

C
$$57.9^o$$

D
$$62^o$$

Solution

Given: A straight rod is partially immersed in water ($$\mu=1.33$$). Its submerged portion appears to be inclined at $$45^o$$ with the surface when viewed vertically from air.
To find the actual inclination of the rod
Solution:
We know,
$$\mu=\dfrac {\sin i}{\sin r}\\\implies 1.33=\dfrac {\sin (45)}{\sin r}\\\implies \sin r=\dfrac {\dfrac 1{\sqrt 2}}{1.33}\\\implies r=\sin^{-1}(0.53)=32.11^\circ$$
Is the angleof refraction.
So the actual inclination is $$90-r=90-32.11=57.9^\circ$$
Question 5
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There is a small black dot at the centre C of a solid glass sphere of refractive index $$\mu$$. When seen from outside, the dot will appear to be located:

A
Away from C for all values of $$\mu$$

B
At C for all values of $$\mu$$

C
At C for $$\mu =1.5$$, but away from C for $$\mu\neq 1.5$$

D
At C only for $$\sqrt{2} \leq \mu \leq 1.5$$

Solution

The object is kept at the centre of the sphere all the rays coming from it are normal to the surface of the sphere. Therefore they will emerge out undecided. The image will appear at the same place away from C for all values of $$\mu$$.
Question 6
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A water drop in air reflects the light rays as

A

B

C

D

Solution

A water drop in air behaves as converging lens.
Question 7
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A microscope is focussed on an ink mark on the top of a table. If we place a glass slab of $$3$$cm thick on it, how should the microscope be moved to focus the ink spot again? The refractive index of glass is $$1.5$$.

A
$$2$$cm upwards

B
$$2$$cm downwards

C
$$1$$cm upwards

D
$$1$$cm downwards

Solution

Given,
The refractive index of the glass is $$n=1.5$$
The thickness of the glass slab is $$ h_o=3$$cm

Refractive index is given as:
$$n=\displaystyle\frac{h_o}{h_i}$$

$$1.5=\displaystyle\frac{3}{h_i}$$

$$\displaystyle h_i=\frac{3}{1.5}=2\ cm$$

So, the microscope needs to be shifted upwards by:
$$h'=3-2=1\ cm$$ upwards.
Question 8
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A ray of light passing through a semicircle block from air is shown below. Consider the given figure and select the correct option which correctly identifies the true (T) and false (F)?
(i) There is no change in the direction of ray at $$P$$ because light ray incident on semicircular block at $$90^{\circ}$$.
(ii) As light rays enters the semicircular block, its frequency remains the same.
(iii) Wavelength of light rays as it enters the block decreases.
(iv) Speed of light as it enters the block increases.

A
T F F T

B
F T T F

C
T T T F

D
F F F T

Solution

The ray of light is normal to the surface of the semicircular block.
The frequency of the light ray remains same.
Wavelength of light rays as it enters the block decreases.
Speed of light as it enters the block decreases because it is entering from rarer to denser medium.
Question 9
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There is a small black dot at the centre $$C$$ of a solid glass sphere of refractive index $$\mu$$. When seen from outside, comment on the location of the dot.

A
away from $$C$$ for all values of $$\mu$$

B
at $$C$$ for all values of $$\mu$$

C
at $$C$$ for $$\mu = 1.5$$, but away from $$C$$ for $$\mu \neq 1.5$$

D
at $$C$$ only for $$\sqrt {2} \leq \mu \leq 1.5$$

Solution

The object is kept at the centre of the sphere all the rays coming from it are normal to the surface of the sphere. Therefore they will emerge out undecided. The image will appear at the same place away from C for all values of $$\mu$$.
Question 10
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What is the required condition, if the light incident on one face of a prism, does not emerge from the other face?

A
$$n < cosec \left (\dfrac {A}{2}\right )$$

B
$$n < \sec \left (\dfrac {A}{2}\right )$$

C
$$n > \sec A$$

D
$$n > cosec \left (\dfrac {A}{2}\right )$$

Solution

The light has to do total internal reflection at the second face so that it cannot come out of the prism.
Condition for total internal reflection $$\sin r > \dfrac{1}{n}$$
But $$A = r+r = 2r$$
$$\implies \ r = \dfrac{A}{2}$$
Thus we get $$\sin \dfrac{A}{2} > \dfrac{1}{n}$$
$$\implies \ n > cosec \dfrac{A}{2}$$