2 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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Question 12 Marks

Obtain the equivalent focal length of combination of thin lenses placed in cantact.
$\left(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\right)$

Answer

→As shown in figure two lenses A and B are arranges so that their principal axis is the same. The focal lengths of these are $f_1$ and $f_2$ respectively. Here we will assume that since both the lenses are thin, their optical centre converge on each other. Let the centre be the point P .
→Let the object be placed at point $O$ beyond the focus of the first lens A . The first lens produces an image at $I_1$. This image $I_1$ serves as a virtual object for the second lens. B producing the final image at I .
→For the image formed by the first lens A,
$\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1}......(1)$
→For the image formed by the second lens B,
$\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}......(2)$
→Adding equations (1) and (2),
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}......(3)$
→If the two lens-system is regarded as equivalent to a single lens of focal length $f$.
$\therefore \quad \frac{1}{v}-\frac{1}{u}=\frac{1}{f}......(4)$
→Comparing equations (3) and (4),
$\therefore \quad \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$
→The derivation is valid for any number of thin lenses, in contact. If several thin lenses of focal length $f_1, f_2, f_3, \ldots$ are in contact, the effective focal length of their combination is given by $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}+\ldots$

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Question 22 Marks

Explain Cassegrain telescpoe with the help of figure.

Answer

→Figure shows construction of cassegrain telescope. It uses two mirrors. Among them, the concave mirror called the object mirror (main mirror) and the convex mirror is called the secondary mirror.
→An object mirror has a larger focal length and a secondary mirror has a less focal length.
→As shown in figure, rays from a distant object are reflected by the objective mirror and focussed at the principal focus.
→But another small convex mirror is placed in the path of these rays. The rays reflected by this mirror are observed through the eyepiece.

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Question 32 Marks

A compound microscope consists of an objective lens of focal length 1 cm , an eyepiece of focal length 2 cm and tube length is 20 cm . Find its magnification.

Answer

→Here, $\begin{aligned} f_0 &=1 cm, \quad f_e&=2 cm \\ L&=20 cm, \quad D&=25 cm \text { (near point) }\end{aligned}$
→magnification in compound microscope,
$\begin{aligned}m & =\frac{ LD }{f_0 f_e} \\m & =\frac{20 \times 25}{1 \times 2} \\\therefore \quad m & =250\end{aligned}$

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Question 42 Marks

Obtain the angular magnification for the image formed at infinity for simple microscope.

Answer

→When an object of height $h$ is placed at the near point as shown in the figure (a), the angle subtended at the eye by the object is $\theta_0$.
→from figure, $\tan \theta_0=\frac{h}{ D }$
for small angle, $\tan \theta_0 \approx \theta_0=\frac{h}{ D }......(1)$
→As shown in figure (b), if an object of height $h$ is placed at the principal focus, its image will be at an infinite distance. This time, the angle subtended at the eye by the object is $\theta_i$.
→from figure, $\tan \theta_i=\frac{h}{f}$
but for small angle, $\tan \theta_i \approx \theta_i=\frac{h}{f}......(2)$
→angular magnification of lens,
$\begin{aligned}m & =\frac{\theta_i}{\theta_0}=\frac{\frac{h}{f}}{\frac{h}{ D }}=\frac{h}{f} \times \frac{ D }{h} \\\therefore m & =\frac{ D }{f}\end{aligned}$
→This is the magnification when the image is at infinity.
→Thus, the magnification obtained with simple microscope is between $\frac{ D }{f}$ to $1+\frac{ D }{f}$.

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Question 52 Marks

Explain simple microscope. Obtain the equation of magnification for the image formed at near point.

Answer

$\rightarrow$ A simple microscope is a converging lens of small focal length.
$\rightarrow$ As shown in figure in order to use such a lens as microscope, the lens is held near the object, one focal length away or less, and eye is positioned close to the lens on the other side.
$\rightarrow$ Here the object is placed in front of the lens in such a way that an erect, magnified and virtual image can be viewed at a near point. $($see figure.$)$
$\rightarrow$ For simple microscope magnification
$m=\frac{v}{u}......(1)$
$\rightarrow$ Applying lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\therefore \frac{1}{u}=\frac{1}{v}-\frac{1}{f}......(2)$
$\rightarrow $ Substituting value of equation $(2)$ in equation $(1),$ magnification $m=v\left(\frac{1}{v}-\frac{1}{f}\right)$
$\therefore m=1-\frac{v}{f}$
$\rightarrow$ but from sign convention we get $v=-D$.
$\therefore m=1+\frac{ D }{f}$

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Question 62 Marks

Write the name of optical instruments.

Answer

→A number of optical devices and instruments have been designed utilising reflecting and refracting properties of mirrors, lenses and prisms. Periscope, Kaleidoscope, binoculars, telescopes, microscopes are some examples of optical devices and instruments that are in common use.
→Human eye is one of the most important optical device the nature has endowed us with.

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Question 72 Marks

Derive $D_m=A\left(n_{21}-1\right)$ for thin lens.

Answer

$\rightarrow$ Refractive index of prism $n_{21}=\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \frac{A}{2}} \ldots(1)$
$\rightarrow$ But for thin prism, prism angle $A$ is small and,
$\therefore \sin \left(\frac{ A + D _m}{2}\right) \approx \frac{ A + D _m}{2}$
$ \sin \left(\frac{ A }{2}\right) \approx \frac{ A }{2}$
$\rightarrow$ Substituting in equation $(1),$
$\therefore n_{21}=\frac{\frac{A+D_m}{2}}{\frac{A}{2}}$
$\therefore n_{21}=\frac{ A + D _m}{A}$
$\therefore n_{21} \cdot A = A + D _m$
$\therefore D _m=n_{21} \cdot A - A$
$\therefore D _m= A \left(n_{21}-1\right)$
$\rightarrow$ From this formula it can be said that thin prism $($Thin prism means $' A\ '$ will be smaller and hence $D _{ m }$ will be smaller, too.$)$ does not deflect light much.

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Question 82 Marks

Obtain the equation of magnification of combination of lenses.

Answer

→Figure shows combination of two lenses A and B. Let their magnification be $m_1$ and $m_2$.
→For lens A, Object distance is $u$ and image distance is $v^{\prime}$. magnification $m_1=\frac{v^{\prime}}{u}......(1)$
→For lens B,
Object distance is $v^{\prime}$ and image distance is $v$.
$\therefore$ magnification $m_2=\frac{\nu}{v^{\prime}}......(2)$
→Suppose magnification for the given lens combination is $m$.
$\therefore$ magnification $m=\frac{v}{u}$
$\therefore m=\frac{v^{\prime}}{u} \times \frac{v}{v^{\prime}}$ (multiplying and dividing by $v^{\prime}$ )
→Substituting values of equation (1) and (2),
$\therefore m=m_1 \times m_2$
→For combination of several lenses, $m=m_1 \times m_2 \times m_3 \times \ldots$.
→Thus, equivalent magnification of combination is an algebraic multiplication of individual magnifications.

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Question 92 Marks

Define power of lens, obtain its equation and write its unit.

Answer

→Power of a lens is a measure of the convergence or divergence, which a lens introduces in the light falling on it.
→Clearly, a lense of shorter focal length bend the incident light more, while converging it in case of a convex lens and diverging it in case of a concave lens.
→"The tangent of the angle by which it converges or diverges a beam of light parallel to the principal axis falling at unit distance from the optical centre is defined as the power P of a lens."

→From figure, $\tan \delta=\frac{h}{f}$;
if $h=1$ then $\tan \delta=\frac{1}{f}$ or $\delta=\frac{1}{f}$
for small value of $\delta$. Thus
power $P =\frac{1}{f}$
→The SI unit for power of a lens is dioptre (D). $\therefore \quad 1 D =1 m^{-1}$.
→Power of a lens is positive for a convex lens and negative for a concave lens.
→When an optician prescribes a corrective lens of power +2.5 D , the required lens is a convex lens of focal length +40 cm . A lens of power of -4.0 D means a concave lens of focal length -25 cm .

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Question 102 Marks

Things to keep in mind to get the image formed by the lens.

Answer

→To find the image of an object by a lens, we can, in principal, take any two rays emanating from a point on an object, trace their paths using the laws of refraction and find the point where the refracted rays meet (or appear to meet). In practice, however, it is convenient to choose any two of the following rays:

(i) A ray emanating from the object parallel to the principle axis of the lens after refraction passes through the second principal focus (in a convex lens) or appears to diverge (in a concave lens) from the first principal focus.
(ii) A ray of light, passing through the optical centre of the lens, emerges without any deviation after refraction.
(iii) A ray of light passing through the first principal focus of a convex lens (or appearing to meet principal axis at second focus point of a concave lens) emerges parallel to the principal axis after refraction.
→Figure (a) and (b) illustrate these rules for a convex and a concave lens.

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Question 112 Marks

What is lens? State their types and draw their diagram.

Answer

→A structure formed by two refracting surfaces and capable of forming an image is called a lens.
→Types of lenses :
(1) Convex lens
(2) Concave lens
(3) Plano convex lens
(4) Plano concave lens
(5) lens formed by meniscus

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Question 122 Marks

Explain the refraction by two transparent spherically curved surface.

Answer

→At infinitesimal part of a spherical surface can be regarded as planar and the same laws of refraction can be applied at every point on the surface.
→Just as for reflection by a spherical mirror, the normal at the point of incidence is perpendicular to the tangent plane to the spherical surface at that point and therefore, passes through its centre of curvature.

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Question 132 Marks

Explain the phenomena of the total internal reflection in right angle prism and the image formed by it.

Answer

→A prism usually has a refractive index of 1.5 giving a corresponding critical angle of $42^{\circ}$.
→As shown in figure (a) a ray of light is deflected by $90^{\circ}$ and thereby produces an image $A ^{\prime} B ^{\prime}$ of an object $A B$.
→As shown in figure (b), a ray of light is deflected by $180^{\circ}$ and an inverted image $A ^{\prime} B ^{\prime}$ is obtained of the object AB .
→As shown in figure (c) such a prism is also used to invert images without changing their size.

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Question 142 Marks

With the help of suitable diagram explain the real depth and apparent depth.

Answer

$\rightarrow$ Generally, we experience that a tank filled with water appears to be raised when viewed vertically or close to vertical.
$\rightarrow$ As shown in figure object $O$ is at the bottom of the tank. When it is viewed from air it observed at $O ^{\prime}$ which is virtual image.
$\rightarrow$ As shown in figure real depth of bottom is $h_2$ and apparent depth is $h_1$.
$\rightarrow$ For viewing near the normal direction, it can be shown that the apparent depth $\left(h_1\right)$ is equal to the ratio of the real depth $\left(h_2\right)$ to the refractive index $(n)$ of the medium.
$\therefore \text { apparent depth }=\frac{\text { real depth }}{\text { refractive index }}$
$\therefore h_1=\frac{h_2}{n}$
$\therefore \frac{h_1}{h_2}=\frac{1}{n}$

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Question 152 Marks

Explain the lateral shift for a ray refracted from a glass slab with parallel sides.

Answer

→As shown in figure ray is incident on a rectangular slab at an angle $i_1$ and is refracted through an angle $r_1$ and travels through the glass slab.
→This ray is incident on the other surface of the slab at an angle $i_2$.
→From this point a ray emerges at an angle $r_2$. Here, $i_1=r_2$ can be easily proved from Snell's law.
→Here, emergent ray is parallel to incident ray. Thus there is no deviation of incident ray.
→The perpendicular distance between the incident ray and the emergent ray is called lateral shift.

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Question 162 Marks

Obtain Primary results from law of refraction.

Answer

$\rightarrow$ Suppose refractive index of two medium are $n_1$ and $n_2$ respectively.
$\rightarrow$ If $n_{21}=$ refractive index of medium $2$ with respect to medium $1$ then,
we get $n_{21}=\frac{n_2}{n_1}$
$\rightarrow$ If $n_{12}=$ refractive index of medium $1$ with respect to medium $2$ then,
we get, $n_{12}=\frac{n_1}{n_2}$
$\therefore n_{12}=\frac{n_1}{n_2}$
$\therefore n_{12}=\frac{1}{\frac{n_2}{n_1}}$
$\therefore n_{12}=\frac{1}{n_{21}}$
$\therefore n_{12} \times n_{21}=1$
$\rightarrow$ If $n_{32}$ is the refractive index of medium $3$ with respect to medium $2$ then,
$n_{32}=n_{31} \times n_{12}$
where, $n_{31}$ is the refractive index of medium $3$ with respect to medium $1$ .
$\therefore n_{32}=\frac{n_3}{n_1} \times \frac{n_1}{n_2}$
$\therefore n_{32}=\frac{n_3}{n_2}$
$\rightarrow$ Suppose five media having different refractive index are given.
$ n_{54} \times n_{43} \times n_{32} \times n_{21}$
$=\frac{n_5}{n_4} \times \frac{n_4}{n_3} \times \frac{n_3}{n_2} \times \frac{n_2}{n_1}$
$\therefore n_{54} \times n_{43} \times n_{32} \times _{21}=\frac{n_5}{n_1}=n_{51}$

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Question 172 Marks

What is optically denser and optically rarer medium?

Answer

$\bullet$ Optical denser medium:

$\rightarrow$ If $n_{21}>1$, then from $\frac{\sin i}{\sin r}=n_{21}$
$ \frac{\sin i}{\sin r}>1$
$\therefore i>r$
$\rightarrow$ Hence, refracted ray bends towards the normal.
$\rightarrow$ Here, medium $2$ is said to be denser as compared to medium $1$ .
$\bullet$Optically rarer medium :

$\rightarrow$ If $n_{21}<1$, then
$ \frac{\sin i}{\sin r}<1$
$\therefore i < r $
$\rightarrow$ Hence, refracted ray goes away from normal.
$\rightarrow$ Here, medium $2$ is said to be rarer as compared to medium $1$ .

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Question 182 Marks

Write Snell's law and derive its general equation.

Answer

$$\rightarrow$$ Snell's Law : For a given two medium the ratio of the sine of the angle of incidence to the sine of angle of refraction is constant.
$\frac{\sin i}{\sin r}=n_{21}......(1)$
where, $n_{21}$ is the refractive index of the second medium with respect to the first medium.

$\rightarrow$ absolute refractive index of medium $(1),$
$n_1=\frac{C}{v_1}$
absolute refractive index of medium $(2),$
$n_2=\frac{ C }{ v _2}$
$($where, $C$ is the velocity of light in vacuum.$)$
$\therefore n_{21}=\frac{v_1}{v_2} \text { and } \frac{n_2}{n_1}=\frac{v_1}{v_2}$
$\therefore n_{21}=\frac{n_2}{n_1}......(2)$
$\rightarrow$ From equation $(1)$ and equation $(2)$,
$\therefore \frac{\sin i}{\sin r}=\frac{n_2}{n_1}$
$\therefore n_1 \sin i=n_2 \sin r$
$\rightarrow$ This is called the general equation of Snell's law.

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Question 192 Marks

What is absolute refractive index? On what factors does it depend ?

Answer

$\rightarrow$ The refractive index of a given medium with respect to vacuum is called its absolute refractive index.
$n =\frac{c}{v}$
$\rightarrow$ It is also equal to the ratio of speed of light in vacuum to speed of light in given medium.
$\rightarrow$ Refractive index is unitless.
$\rightarrow$ Dimensional formula: $M^0L^0T^0$
$\rightarrow$ Value of refractive index of any medium depends on type of medium, temperature and wavelength of light.
$\rightarrow$ The refractive index of a medium and the wavelength of light in the same medium are inversely proportional to each other. $\left(n \propto \frac{1}{\lambda}\right)$

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Question 202 Marks

What is refraction of light? Explain law of refraction.

Answer

→"The direction of propagation of an obliquely incident ray of light that enters the other medium, changes at the interface of the two medium. This phenomenon is called refraction of light."

→Laws of refraction:
(i) The incident ray, the refracted ray and normal to the interface at the point of incidence, all lie in the same plane.
(ii) For the given two medium the ratio of the sine of the angle of incidence to the sine of angle of refraction is constant. Let the angle of incidence i and the angle of refraction r be the angles made with the normal to the ray of incidence and ray of refraction respectively.
$\frac{\sin i}{\sin r}=n_{21}$<br>where, $n_{21}$ is the refractive index of the second medium with respect to the first medium.<br>Equation (1) is the Snell's law.

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Question 212 Marks

What is linear magnification? Obtain equation of linear magnification for mirror.

Answer

$\rightarrow$ Linear magnification is the ratio of the height of the image $\left(h^{\prime}\right)$ to the height of the object $(h)$.
$\therefore m=\frac{h^{\prime}}{h}$

$\rightarrow$ As shown in figure object $AB$ is placed slight far from centre of curvature. Its image is found between $C$ and $F$.
$\rightarrow$ Here height of the object is $h$ and height of image is $h^{\prime}$.
$\rightarrow$ As per figure $\triangle ABP$ and $\triangle A ^{\prime} B ^{\prime} P$ are similar triangle.
$\therefore \frac{ B ^{\prime} A ^{\prime}}{ BA }=\frac{ B ^{\prime} P }{ BP }$
$\rightarrow$ According to sign convention,
$ B ^{\prime} A ^{\prime}=-h^{\prime}, BA =h$
$B^{\prime} P =- v , BP =-u$
$\therefore \frac{-h^{\prime}}{h}=\frac{- v }{-u}$
$\therefore \frac{h^{\prime}}{h}=-\frac{v}{u}$
$\rightarrow$ From equation $(1),$
$m=-\frac{v}{u}$

No.	Type of mirror	Image type and size	Magnification
$1$	Plane	Virtual, erect and same as of object	$+1$
$2$	Concave	Real, inverted and same as of object	$-1$
$3$	Concave	Real, inverted and magnified	$> 1$ and negative
$4$	Concave	Real, inverted and diminished	$< 1$ and negative
$5$	Concave	Virtual, erect and magnified	$> 1$ and negative
$6$	Convex	Virtual, erect and diminished	$< 1$ and positive

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Question 222 Marks

Explain the rules for obtaining image due to reflection from a spherical mirror.

Answer

→We can take any two rays emanating from a point on an object, trace their paths, find their point of intersection and thus obtain the image of the point due to reflection at a spherical mirror.

→In practice, however it is convenient to choose any two of the following rays:
→(i) Rays incident parallel to the principal axis are reflected and goes through the focus of the mirror.
(ii) The ray passing through the centre of curvature of a concave mirror or appearing to pass through it for a convex mirror. The reflected ray simply retraces the path.
(iii) The ray passing through the focus of the concave mirror or appearing to pass through the focus of a convex mirror. The reflected ray is parallel to the principal axis.
(iv) The ray incident at any angle at the pole (P). The reflected ray follows laws of reflection.

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Question 232 Marks

What is image? Explain its types.

Answer

→If rays emanating from a point actually meet at another point after reflection and/or refraction, that point is called the image of the first point.
→There are two types of image:
(1) Real image : If the reflected rays actually converge at any point, then image is called the real image.
(2) Virtual image: If the reflected rays do not actually meet but appear to diverge from the point when produced backward then image is called virtual image.

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Question 242 Marks

Derive R=2f for spherical mirror.
(Where, R= radius of curvature and f = focal length)###Obtain relation between radius of curvature and focal length of spherical mirror.

Answer

→As shown in figure, C is center of curvature of the mirror and F is the principal focus of the mirror.
→As shown in figure, a ray is incident at an angle $\theta$ near a point M on the mirror. CM is the perpendicular drawn to the surface. MD is the perpendicular drawn from point M on the principal axis.
→From figure, $\angle MCP =\theta$ and $\angle MFP =2 \theta$
→Now, $\tan \theta=\frac{ MD }{ CD }......(1)$
and $\tan 2 \theta=\frac{ MD }{ FD }......(2)$
→Since $\theta$ is extremely small for paraxial rays $\tan \theta \approx \theta$ and $\tan 2 \theta \approx 2 \theta$
→From equation (1) and equation (2),
$\begin{aligned}& \theta=\frac{M D}{C D} \text { and } 2 \theta=\frac{M D}{F D} \\\therefore \quad & 2\left(\frac{M D}{C D}\right)=\frac{M D}{F D} \\\therefore \quad & \frac{2}{C D}=\frac{1}{F D} \\\therefore \quad & C D=2 F D......(3)\end{aligned}$
→Now for small $\theta$ the point D is very close to the point $P$.
→Therefore, $FD = FP =f$ and $CD = CP = R$
Substituting above values in equation (3),
$\begin{aligned}R & =2 f \\\text { or } f & =\frac{ R }{2}\end{aligned}$

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Question 252 Marks

Discuss sign convention of distance for reflection by spherical mirror and refraction by spherical lens.

Answer

(i) According to the Cartesian sign method all distances are measured from the pole for the mirror. Whereas for lens all the distances are measured from its optical centre.

(ii) The distances measured in the same direction as the incident light are taken as positive and those measured in the direction opposite to the direction of incident light are taken as negative.
(iii) The height measured upward with respect to X-axis and normal to the principal axis of the mirror/lens are taken as positive. The heights measured downwards are taken as negative.
→With a common accepted convention, it turns out that a single formula for spherical mirrors and a single formula for spherical lenses can handle all different cases.

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Question 262 Marks

Definitions needed to understand reflection from a Spherical mirror :

Answer

→(1) Centre of Curvature (C): The centre of the spherical shell from which the mirror is made, is called the centre of curvature (C) of the mirror.
(2) Radius of Curvature (R): The radius of the spherical shell from which the mirror is made is called the radius of curvature (R).
(3) Pole (P): The centre of the reflecting surface of a mirror is called pole (P).
(4) Principal Axis: The imaginary line passing through the pole and the centre of curvature is called the principal axis of the mirror.
(5) Aperture: The diameter of the reflecting surface is called aperture of the mirror (QQ').
(6) Principal Focus (F): The point where the rays parallel to the principal axis meet (concave mirror), or appear to meet (convex mirror), after reflection is called the principal focus of the mirror.

(7) Focal Plane: A plane passing through the principal focus and normal to the principal axis is called the focal plane of the mirror.
(8) Focal Length (f): This distance between the pole and the principal focus of a mirror is called focal length.
(9) Paraxial Rays: Rays close to the principal axis are called paraxial rays.

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Question 272 Marks

Write laws of reflection.

Answer

→(i) The angle of incidence and the angle of reflection are same. (i = r)
(ii) The incident ray, reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane. This plane is called the plane of incidence.
(iii) The incident ray and reflected ray lie on opposite sides of the normal.

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Question 282 Marks

What is reflection of light? Define angle of incidence and angle of reflection.

Answer

→The phenomenon in which the light incident on a reflective surface returns back from that surface is called reflection of light.
→Angle of incidence (i): The angle between incident ray and the normal to the reflecting surface at the point of incidence is called angle of incidence.
→Angle of reflection (r): The angle between reflected ray and the normal to the reflecting surface.

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Question 292 Marks

Explain ray of light and beam of light.

Answer

→A light wave can be considered to travel from one point to another, along a straight line joining them. This path is called a ray of light. A bundle of such rays constitutes a beam of light.

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Question 302 Marks

Briefly explain about light.

Answer

$\rightarrow$ The human eye is only sensitive to small range of electromagnetic spectrum.
$\rightarrow$ This region is known as visible light.
$\rightarrow$ The wavelength of this region is $400 \ nm$ to $750 \ nm.$
$\rightarrow$ Two important points about light:
$(i)$ Light travels with enormous speed. Speed of light is limited and measurable.
$\rightarrow$ Its presently accepted value in Vacuum is $c = 2.99792458 \times 10^8 m/s$
$\rightarrow$ For ease of calculations the velocity of light in free space is taken as $c = 3 \times 10^8 m/s$
$(ii)$ Light travels in a straight line.

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