3 Marks Question

Question 13 Marks

Obtain the relationship between object distance $(u),$ image distance $(v)$ and focal length $(f)$ for a concave mirror. Obtain $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ formula for a concave mirror.

Answer

$\rightarrow$ A mirror with small aperture is shown in figure. An object $AB$ is placed in front of the mirror at some distance from the centre of curvature.
$\rightarrow$ Three rays emanating from $A$ are reflected by a mirror and converge at point $A\ ^{\prime}$.
So the image of the object $A B$ is given by $A\ ^{\prime} B\ ^{\prime}$ between $C$ and $F$ .
$\rightarrow$ From figure the two right$-$angled triangles $\triangle A\ ^{\prime} B\ ^{\prime} F$ and $\triangle MPF$ are similar. $($For paraxial rays$, MP$ can be considered to be a straight line perpendicular to $CP.)$
$\rightarrow$ Therefore, $\frac{ A ^{\prime} B ^{\prime}}{ MP }=\frac{ B ^{\prime} F }{ FP }$
But, $AB = MP$$\frac{ A ^{\prime} B ^{\prime}}{ AB }=\frac{ B ^{\prime} F }{ FP }......(1)$
$\rightarrow$ The right angled triangles $\triangle ABP$ and $\triangle A ^{\prime} B ^{\prime} P$ are similar.
Therefore, $\frac{A^{\prime} B^{\prime}}{A B}=\frac{B^{\prime} P}{B P}......(2)$
$\rightarrow$ Comparing equations $(1)$ and $(2),$
We get,
$\therefore \frac{B^{\prime} F }{ FP }=\frac{ B ^{\prime} P }{ BP }$
$\rightarrow$ But, $B ^{\prime} F = PB ^{\prime}- FP$
$\therefore \frac{ PB ^{\prime}- FP }{ FP }=\frac{ B ^{\prime} P }{ BP }......(3)$
But$ , B ^{\prime} P =- v , FP =-f, BP =-u$
$($according to sign convention all three have negative signs$)$
$\rightarrow$ using these in equation $(3),$ we get
$\frac{-v+f}{-f}=\frac{-v}{-u}$
$\therefore \frac{-v}{-f}-\frac{f}{f}=\frac{v}{u}$
$\therefore \frac{v}{f}-1=\frac{v}{u}$
$\rightarrow$ Now dividing by $v$,
$\therefore \frac{v}{f v}-\frac{1}{v}=\frac{v}{u v}$
$\therefore \frac{1}{f}-\frac{1}{v}=\frac{1}{u}$
$\therefore \frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\rightarrow$ It is called mirror equation.

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Question 23 Marks

Explain the experiment which represents the total internal reflection.

Answer

→Take a glass beaker with clear water in it. Add a few drops of milk or any other suspension to water and stir the water a few times, so that it becomes a little turbid. Take a laser pointer and shine its beam through the turbid water. You will find that the path of the beam inside the water shines brightly.
→As shown in fig (a) shine the beam from below the beaker such that it strikes at the upper water surface at the other end. Partial reflection and partial refraction of an incident ray can be observed in water.
→As shown in figure (b) direct the laser beam from one side of the beaker such that it strikes the upper surface of water more obliquely. Adjust the direction of laser beam until you find the angle, for which the refraction above the water surface is totally absent and the incident ray undergoes total internal reflection in water itself.
→Pour this water in a long test tube and shine the laser light from top as shown in fig (c). Adjust the direction of the laser beam such that it is totally internally reflected every time it strikes the walls of the tube.

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Question 33 Marks

Explain internal reflection and total internal reflection and also derive the necessary formula for it.

Answer

→When a ray of light travels from an optically denser medium to a rarer medium at the interface, it is partly reflected back into the same medium and partly refracted to the second medium. This reflection is called the internal reflection.
→When a ray of light enters from a denser medium to a rarer medium it bends away from the normal.

→$AO _1$ is the incident ray to ray $AO _1 B . O _1 C$ is partially reflected and $O _1 B$ is the partially refracted ray. Here, the ray passes from denser medium to the rarer medium, so incidence angle $(i) <$ refraction angle $(r)$.
→As we move from $O _1$ to $O _2, O _3 \ldots$. the value of incidence angle increases, As a result the value of refraction angle also increases.
→The angle of refraction is $90^{\circ}$ for the ray $AO _3$.
→Here the refracted ray becomes parallel to the surface separating the two media. This is shown in figure by the ray $AO _3 D$.
→The angle of incidence for which the refraction angle becomes $\frac{\pi}{2}$, that incidence angle is called critical angle.
→If the angle of incidence is increased still further (ray $AO _4$ ) refraction is not possible and the incident ray is totally reflected. This is called total internal reflection.
→According to Snell's law at point $O _3$,
$\begin{aligned}& n_1 \sin i_{ C }=n_2 \sin 90^{\circ} \\\therefore & \sin i_{ C }=\frac{n_2(1)}{n_1} \\\therefore & \sin i_{ C }=\frac{n_2}{n_1}=n_{21}\end{aligned}$
(where, $n_{21}$ is refractive index of medium 2 with respect to medium 1)
→If $n_2=1$ (air) and $n_1=n$, then
$\therefore \sin i_{ C }=\frac{1}{n}$

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Question 43 Marks

Explain the construction of compound microscope with the help of figure and derive the equation of magnification.

Answer

$\rightarrow $ The maximum magnification available with a simple microscope is $9$. For much larger magnifications one uses two lenses, one compounding the effect of the other. This is known as a compound microscope.

$\rightarrow $A schematic diagram of a compound microscope is shown in figure. The lens nearest the object is called the objective. It's focal length is $f_0$. The lens nearest the observer, called the eyepiece. It's focal length is $f_{e^.}$. For compound microscope $f_0$
$\rightarrow $An objective produces real, inverted and magnified image of the object. This serves as the object for the eyepiece. Eyepiece functions essentially like a simple microscope and produces final image, which is enlarged and virtual.
$\rightarrow $For an objective lens of compound microscope.
$m_0=\frac{h^{\prime}}{h}......(1)$
$($from definition$)$
Where, $h^{\prime}=$ height of image
$h=$ height of object
$\rightarrow $from figure,
$\begin{array}{l}\tan \beta=\frac{ AB }{ OB _1} \text { and } \tan \beta=\frac{ A ^{\prime} B ^{\prime}}{ B _1 B^{\prime}} \\\therefore \quad \frac{ AB }{ OB _1}=\frac{ A ^{\prime} B ^{\prime}}{ B _1 B^{\prime}} \\\therefore \quad \frac{ A ^{\prime} B ^{\prime}}{ AB }=\frac{ B _1 B^{\prime}}{ OB _1} \\\end{array}$
$\rightarrow$ But $A ^{\prime} B ^{\prime}=h^{\prime}, AB =h, B_1 B^{\prime}= L , OB _1=f_0 \\\therefore \frac{h^{\prime}}{h}=\frac{ L }{f_0} \\\therefore m_0=\frac{ L }{f_0}......(2)$
Here $L$ is tube length.
$\rightarrow $Tube length $=$ "the distance between the second focal point of the objective and the first focal point of the eyepiece is called the tube length."
$\rightarrow $Here, the eyepiece becomes a simple microscope. As a result when the final image is formed at the near point, the magnification is
$m_e=\left[1+\frac{ D }{f_e}\right]$
$\rightarrow $When the final image is formed at infinity, the magnification.
$m_e=\frac{ D }{f_e}......(3)$
$\rightarrow $Thus, the total magnification, when the final image is formed at infinity, is
$m=m_0 \times m_e=\frac{ L }{f_0} \times \frac{ D }{f_e}$

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Question 53 Marks

Using lens maker's formula obtain the equation of lens.

Answer

→For refraction at two spherical surfaces of lens,
$\begin{aligned}& \frac{n_1}{ OB }+\frac{n_1}{ DI }=\left(n_2-n_1\right)\left(\frac{1}{ BC _1}+\frac{1}{ DC _2}\right) \\\therefore \quad & n_1\left(\frac{1}{ OB }+\frac{1}{ DI }\right)=\left(n_2-n_1\right)\left(\frac{1}{ BC _1}+\frac{1}{ DC _2}\right) \\\therefore \quad & \frac{1}{ OB }+\frac{1}{ DI }=\left(n_{21}-1\right)\left(\frac{1}{ BC _1}+\frac{1}{ DC _2}\right)\end{aligned}$
→using sign convention,
$OB =-u, DI =v, BC _1= R _1 \text { and } DC _2=- R _2.$
Substituting these values in above equation,
$\therefore-\frac{1}{u}+\frac{1}{v}=\left(n_{21}-1\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$ ...(1)
→lens maker's formula,
$\therefore \quad \frac{1}{f}=\left(n_{21}-1\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$ ...(2)
→Comparing equation (1) and (2),
$\therefore \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
→This equation is known as lens equation.
→This equation is true for both convex and concave lens.

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Question 63 Marks

Explain two main important points for astronomical telescope.

Answer

→The main considerations with an astronomical telescope are its light gathering power and its resolution (or resolving power).
→In objective lens light gathering power depends on the area of the objective.
→With larger diameters, fainter objects can be observed. The resolving power or the ability to observe two objects distinctly also depends on the diameter of the objective. So the desirable aim in optical telescope is to make them with objective of large diameter.
→The largest lens objective in use has a diameter of 40 inch (~ 1.02 m). It is at the Yerks observatory in Wisconsin, USA.

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Question 73 Marks

What are reflecting type of telescope ? Explain the advantages and disadvantages of reflecting telescopes as compared to refracting telescope and also give the solution.

Answer

→A chromatic aberration occurs when an image is formed in a refracting telescope. Due to which the image is not clearly visible.
→A modern telescope that uses a concave mirror rather than a lens for the objective is called a reflecting telescope.

Advantages:
(1) There is no chromatic aberration in a mirror.
(2) If a parabolic reflecting surface is chosen, spherical aberration is also removed.
(3) A mirror weighs much less than a lens of equivalent optical quality.
(4) It can be easily supported.
Disadvantages:

→The objective mirror focuses light inside the telescope tube. One must have an eyepiece and the observer right there obstructing some light.
→To solve this problem a viewer sits near the focal point of the mirror in a small cage. Another solution to the problem is to deflect the light being focused by another mirror.

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Question 83 Marks

Explain the construction of refracting telescope by figure and obtain the equation of magnification.

Answer

$\rightarrow $ Figure shows a refracting telescope.
The telescope is used to observe distant objects.
$\rightarrow$ As shown in figure in this telescope two lenses are used.
The lens facing the object is called objective. It's focal length is $f_0$.
The lens near the eye is called eyepiece. It's focal length is $f_i$. Here $f_0>f_e$.
$\rightarrow $ Light from a distant object enters the objective and a real image is formed in the tube at its second focal point.
$\rightarrow$ This image is the object for the eye $-$ piece.
The eye piece magnifies this image producing a final inverted image.
$\rightarrow$ The ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\alpha$ which the object subtends at the lens or the eye is called the magnifying power $m$.
$\therefore m=\frac{\beta}{\alpha}......(1)$
$\rightarrow $ From figure, $\tan \alpha \approx \alpha=\frac{ A ^{\prime} B ^{\prime}}{ OB ^{\prime}}=\frac{h}{f_0}......(2)$
$\tan \beta \approx \beta=\frac{ A ^{\prime} B ^{\prime}}{ B ^{\prime} E }=\frac{h}{f_e}......(3)$
$\rightarrow$ Substituting values of equation $(2)$ and equation
$(3)$ in equation $(1)$.
$\therefore m=\frac{h}{f_e} \times \frac{f_0}{h}$
$\therefore m=\frac{f_0}{f_e}$
$\rightarrow$ This equation shows that to increase the magnification of the telescope, focal length of the objective should be increased and focal length of the eyepiece should be reduced and the length of tube should be $L >f_0+f_e$.
$\rightarrow f_0+f_e$ is the tube length or optical length of the telescope.

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Question 93 Marks

Explain principle, construction and working of optical fibre.

Answer

→Principle : total internal reflection
→Fabricated from : composite glass/quartz fibres. (diameter 10 to $100 \mu m$ )

→Construction : Figure shows construction of optical fibre. Each fibre consists of certain core and certain cladding.
→The refractive index of the material of the core is higher than that of the cladding, which increases the probability of total internal reflection of the light ray.
→Function : Optical fiber is designed in such a way that the light reflected at one side of inner surface striked the other at an angle larger than the critical angle. As a result, light repeatedly undergoes total internal reflection along the length of the fiber and eventually exits at the other end.
→Since light undergoes total internal reflection at each stage, there is no appreciable loss in the intensity of the light signal. Even if the fibre is bent, light can easily travel along its length.
$\bullet$ Uses :
→Electrical signals are converted into light with the help of transducers. Optical fibres are extensively used for tansmitting and receiving them.
→Optical fibres can also be used for transmission of optical signals.
→For example, these are used as light pipe to facilitate visual examination of internal organs like oesophagus, stomach and intestines.
→Decorative lamp in which the free ends of plastic fibres form a fountain like structure. Its one end is free and the other end is fixed over an electric lamp. When the lamp is switched on, the light travels from the bottom of each fibre and appears at the tip of its free end as a dot of light. The fibres in such decorative lamps are optical fibres.
→The main requirement in fabricating optical fibre : Light travels for long distances inside them. However, it absorbs very little light. That is why optical fibres are made from materials like quartz.
→In silica glass fibres it is possible to transmit more than $95 \%$ of the light over a fibre length of 1 km .

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Question 103 Marks

Obtain the equation of equivalent power for combination of thin lenses placed in contact.

Answer

→Equivalent focal length of combination of two lenses A and B as shown in figure,
$\begin{aligned}\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}......(1) & \\\text { Where, } f_1 & =\text { focal length of lens } A \\f_2 & =\text { focal length of lens } B\end{aligned}$
→Let the power of lenses $A$ and $B$ be $P_1$ and $P_2$ respectively.
$\therefore \quad P_1=\frac{1}{f_1} \text { and } P_2=\frac{1}{f_2}$
→Let the equivalent power of combination is $P$.
$\therefore P =\frac{1}{f}$
→from equation (1) we get $P=P_1+P_2$.
→Equivalent power of several lens combinations,
$P = P _1+ P _2+ P _3+\ldots$
→Equivalent power of combination is an algebraic sum of individual powers.

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Question 113 Marks

Obtain the magnification by lens after giving the definition.

Answer

$\rightarrow$ The ratio of the size of the image obtained by lens to that of the object is called magnification.
$m=\frac{h^{\prime}}{h}......(1)$
$\rightarrow$ As shown in figure $AB$ is an object and its image is $A ^{\prime} B ^{\prime}$.
$\rightarrow$ From figure $\triangle ABP$ and $\triangle A ^{\prime} B ^{\prime} P$ are similar triangles.
$\therefore \frac{ A ^{\prime} B ^{\prime}}{ AB }=\frac{ B ^{\prime} P }{ BP }$
$\rightarrow$ But $, A ^{\prime} B ^{\prime}=-h^{\prime} AB =h$
$ B^{\prime} P =v BP =-u$
$\rightarrow$ Substituting these values in above equation,
$\therefore \frac{-h^{\prime}}{h}=-\frac{v}{u}$
$\therefore \frac{h^{\prime}}{h}=\frac{v}{u}......(2)$
$\rightarrow$ Comparing equation $(1) $ and $(2),$
$\therefore m=\frac{v}{u}$
$\rightarrow$ If the image formed by lens is virtual then magnification $(m)$ is positive and if the image formed by lens is real then magnification $(m)$ is negative.

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