Ray Optics and Optical Test

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Ray Optics and Optical Test - 43

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Question 1
1 / -0

For which of the two lenses $$L_1$$ and $$L_2$$ the chromatic aberration will be more if the radii of curvature of the surfaces of $$L_1$$ are less than those of $$L_2$$:

A
$$L_1$$

B
$$L_2$$

C
$$L_1=L_2$$

D
none of the above

Solution

We know that the focal length of a lens $$f\propto R$$ where R=radius of curvature of surfaces of lens

And, chromatic aberration $$\propto f$$

Hence, larger the radius more will be chromatic aberration.

And, radius of curvature of surface is more for $$L_2$$.

Hence chromatic aberration will be more for $$L_2$$

Answer-(B)
Question 2
1 / -0

White light is incident normally on a glass slab. Inside the glass slab,

A
Red light travels faster than other colours

B
Violet light travels faster than other colours

C
Yellow light travels faster than other colours

D
All colours travel with the same speed

Solution

In media like glass, different wavelengths in the electromagnetic spectrum travel at different speeds.
Refractive index ($$\mu$$) is related to wavelength ($$\lambda$$) of light as
\begin{equation}
\mu=A+\frac{B}{\lambda^2}
\end{equation}
Again $$\mu$$ = Speed of light in Vaccum/Speed of light in medium
Hence, speed of light in media is proportional to wavelength...
Since wavelength increases from Violet to Red, Red travels the fastest.
Question 3
1 / -0

A luminous object is separated from a screen by distance $$d$$. A convex lens is placed between the object and the screen such that it forms a distinct image on the screen. The maximum possible focal length of this convex lens is

A
$$4d$$

B
$$2d$$

C
$$d/2$$

D
$$d/4$$

Solution

Distance between object and the screen is $$d$$.

Let the distance between the object and the lens be $$x$$ i.e. object distance $$u = -x$$

Thus image distance $$v = d-x$$

From lens formula, $$\dfrac{1}{v}-\dfrac{1}{u} =\dfrac{1}{f}$$

$$\therefore$$ $$\dfrac{1}{d-x}-\dfrac{1}{-x} =\dfrac{1}{f}$$ $$\implies f = \dfrac{xd-x^2}{d}$$

For maximum focal length, $$\dfrac{df}{dx} = 0$$

Thus $$\dfrac{d}{dx}(\dfrac{xd-x^2}{d}) = 0$$

We get , $$d-2x=0$$ $$\implies x = \dfrac{d}{2}$$

Hence focal length is maximum when $$x = d/2$$.

Maximum focal length $$f = \dfrac{d(\dfrac{d}{2}) - \dfrac{d^2}{4}}{d} =\dfrac{d}{4}$$
Question 4
1 / -0

Find the refractive index of the glass which is a symmetrical convergent lens if its focal length is equal to the radius of curvature of its surface

A
2

B
1.5

C
1

D
$$\dfrac{4}{3}$$

Solution
Question 5
1 / -0

The distance between principal focus and optical center of the lens is ________.

A
Diameter

B
Focal length

C
Principal axis

D
Optical center

Solution

As shown in the figure, the focal length is the distance between the principal focus and the optical center of the lens.
Question 6
1 / -0

A vessel of depth t is half filled with a liquid having refractive index $$n_1$$ and the other half is filled with water of having refractive index $$n_2$$. The apparent depth of the vessel as viewed from top is:

A
$$\dfrac{2t (n_1 + n_2)}{n_1 n_2}$$

B
$$\dfrac{t (n_1 n_2)}{(n_1 + n_2)}$$

C
$$\dfrac{t (n_1 + n_2)}{2n_1 n_2}$$

D
$$\dfrac{ n_1 n_2}{(n_1 + n_2)t}$$

Solution

Let apparent depth be $$d$$.

Refer the ray diagram in the attached figure.
$$\tan \theta_1 = \cfrac{x}{t/2} ...........(i)$$
$$\tan \theta_2 = \cfrac{y}{t/2} ..........(ii)$$
$$\tan \theta = \cfrac{x+y}{d} .............(iii)$$

By Snell's Law,
$$n_1 \sin \theta_1 = n_2\sin \theta_2 = \sin \theta $$
For small angles, $$\sin \theta \approx \tan \theta$$
$$\sin \theta_1 \approx \tan \theta_1$$
$$\sin \theta_2 \approx \tan \theta_2$$

Hence, $$\tan \theta = n_1 \tan \theta_1 ...............(iv)$$
$$\tan \theta = n_2 \tan \theta_2 .............(v)$$

Substituting (i), (ii) and (iii) in (iv) and (v), we get
$$\cfrac{x+y}{d} = n_1\cfrac{2x}{t} ...........(vi)$$
$$\cfrac{x+y}{d} = n_2\cfrac{2y}{t} ...........(vii)$$

Eliminating $$x$$ and $$y$$ from (vi) and (vii), we get
$$d = \cfrac{t(n_1+n_2)}{2n_1n_2}$$
Question 7
1 / -0

According to the new cartesian sign convention, the ________ is taken as origin

A
Centre of curvature

B
Pole of the mirror

C
Focus

D
Any point on the principal axis

Solution

According to the new cartesian sign convention, pole of the mirror is taken as the origin. All the measurements of object distance, image distance and the focal length are recorded from the pole of mirror. All measurements are positive in the direction of incident rays whereas that and its opposite measurements are taken as negative.
Question 8
1 / -0

Which coloured light scatters least by fog and smoke?

A
Green

B
Yellow

C
Blue

D
Red

Solution

The color of the scattered light depends on the size of the scattering particles. The molecules of air and other fine particles in the atmosphere have sizes smaller than the wavelength of visible light. Red color light has a maximum wavelength and hence is scattered the least.
Question 9
1 / -0

Which colour of light is scattered the most (maximum) in the atmosphere?

A
Red

B
Yellow

C
Blue

D
Green

Solution

The molecules of air and other fine particles in the atmosphere have their size smaller than the wavelength of visible light. Thus, they are more effective in scattering light of shorter wavelengths, e.g. blue light. The red light has a larger wavelength than blue light. Thus, when sunlight is passing through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelengths) more strongly than red.
Question 10
1 / -0

A quarter cylinder of radius $$R$$ and refractive index $$1.5$$ is placed on a table. A point object $$P$$ is kept at a distance of $$mR$$ from it as shown in figure. For what value of $$m$$ for which a ray from $$P$$ will emerge parallel to the table?

A
$$\dfrac { 2 }{ 3 } $$

B
$$\dfrac { 3 }{ 2 } $$

C
$$\dfrac { 4 }{ 3 } $$

D
$$\dfrac { 3 }{ 4 } $$

Solution

Refraction at plane surface (1),

$${ n }_{ 1 }=1$$, $${ n }_{ 2 }=1.5=\dfrac { 3 }{ 2 }$$, $$u=-mR$$, $$v=$$?

Using formula, $$\dfrac { { n }_{ 2 } }{ v } -\dfrac { { n }_{ 1 } }{ u } =\dfrac { { n }_{ 2 }-{ n }_{ 1 } }{ R }$$

$$ \because R=\infty$$ for plane surface.

$$ \because \dfrac { { 3 }/{ 2 } }{ v } -\dfrac { 1 }{ \left( -mR \right) } =0$$

$$\therefore v=-\dfrac { 3 }{ 2 } mR$$

$$u=-\left( \dfrac { 3 }{ 2 } mR+R \right)$$ and $$ R=-R$$, $$v=\infty$$, $${ n }_{ 1 }=\dfrac { 3 }{ 2 }$$, $${ n }_{ 2 }=1$$

$$\dfrac { 1 }{ \infty } -\dfrac { { 3 }/{ 2 } }{ \left[ -\left( \dfrac { 3 }{ 2 } mR+R \right) \right] } =\dfrac { 1-{ 3 }/{ 2 } }{ -R }$$

$$ \Rightarrow \dfrac { { 3 }/{ 2 } }{ \dfrac { R\left( 3m+2 \right) }{ 2 } } =\dfrac { { 1 }/{ 2 } }{ R }$$

$$ \Rightarrow \dfrac { 6 }{ 3m+2 } =1\Rightarrow 3m+2=6$$

$$m=\dfrac { 4 }{ 3 } $$