PhysicsKund Physics class 9 to 12 NCERT, JEE - NEET

Definition of Refraction of light : The change in direction of light when it passes from one medium to another obliquely is called refraction of light . In other words , the bending of light when it goes from one medium to another obliquely is called Refraction of light. Important terms Related to Refraction: The plane surface that refracts light is known as the surface of refraction. Point of incidence: The point on the surface of refraction , where the ray of light is incident is known as the point of incidence . "O" is the point of incidence. Incident Ray : The ray of light which strikes the surface of refraction at the point of incidence , is known as the incident Ray . AO is the incident Ray. Refracted ray : The ray which travels from the point of incidence to other medium is known as the refracted ray. OB is the refracted ray. Normal : Perpendicular drawn on the surface of refraction at the point of incidence is called normal. NON' is the normal on the surface PQ. A...

Combination of Thin Lenses in Contact Equivalent Focal length, Power and Magnification  (i) Equivalent focal length of the combination of Lenses :  Consider two thin lenses of focal lengths $f_1$ and $f_2$ respectively, placed in contact with each other . Let $O$ be the point object placed on the principal axis of the lenses in contact. If the first lens forms an image $I_1$ of the object $O$ at a distance $v_1$ from it. $\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} \quad \text{(Lens formula)} \quad \dots (i)$ Since the second lens is in contact with the first, so $I_1$ acts as a virtual object for the second lens which forms the image $I$ at a distance $v$ from it. $\frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2} \quad \dots (ii)$ Adding eqs. (i) and (ii), we get $\frac{1}{v_1} - \frac{1}{u} +\frac{1}{v} - \frac{1}{v_1}$ $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2}$ or $\frac{1}{F} = $\frac{1}{v} - \frac{1}{u} \quad \dots (iii)$ Where $\frac{1}{F} = \frac{1}{f_1} + \...

Show that the bottom of a water tank appears to be raised. Hence find an expression for the normal shift in the position of an object placed in a denser medium. Consider a tank filled with water upto the level PQ. Let an object O (say a coin) lies at the bottom of the water tank. The depth AO = $t$ of the object is known as the real depth. the object appears at position I instead of O when viewed obliquely. The depth AI is known as apparent depth of the object O.  According to Snell's law, ${ }^{w}n_{a} = \frac{\sin i}{\sin r} \quad \dots (i)$ From $\Delta AOC$,  $\sin i = \frac{AC}{OC}$ and  from $\Delta AIC$,  $\sin r = \frac{AC}{IC}$ Substituting the values of $\sin i$ and $\sin r$ in equation (i), we get ${ }^{w}n_{a} = \frac{\frac{AC}{OC}}{\frac{AC}{IC}}$ ${ }^{w}n_{a} = \frac{AC}{OC} \times \frac{IC}{AC} = \frac{IC}{OC}$ Since point C lies very close to A, so IC $\approx$ AI and OC $\approx$ AO ${ }^{w}n_{a} = \frac{AI}{AO}$ ${ }^{a}n_{w} = \frac{1}{{ }^{w}n_{a...

Notes : Class 12 Physics Chapter 9 Ray Optics and Optical Instruments - Physics Kund Definition and Law of Reflection Sign Convention for Reflection of Light Derive Relation Between Focal Length and Radius of Curvature Derive Mirror Formula (Equation) and Magnification for Convex Mirror Derive Mirror Formula and Magnification for Concave Mirror – Real and Virtual Image Refraction of Light Principle of Reversibility Refraction of Light Through a Glass Slab Real Depth and Apparent Depth Total Internal Reflection and Optical Fibres Refraction at a Spherical Surface Derivation of Lens Maker's Formula and Refraction by a Lens Power of a Lens Combination of Thin Lenses in Contact Refraction of Light Through a Prism Simple Microscope and Its Magnifying Power Optical Instruments: Compound Microscope and Its Magnifying Power Astronomical Telescope and Its Magnifying Power

What is a simple microscope ? With the help of a diagram, explain the working of a simple microscope. Also write the uses of simple microscope. A simple microscope is a converging lens of small focal length to see very small objects as magnified one. It consists of a convex lens of small focal length. A magnifying glass is an example of a simple microscope. Principle : A simple microscope is based upon the fact that an object placed between the optical centre and the focus of a convex lens, forms a virtual, erect and magnified image on the same side of the lens. The image is formed at the least distance of distinct vision (i.e. 25 cm) from the eye. Working : Consider a convex lens of focal length $f$. Let AB be an object which lies between the optical centre (C) and the focus (F) of the lens. The rays of light from the object do not meet after refraction through the lens. They appear to come from a point B' so that A'B' is the virtual image of the object AB. This image is e...

Topic : Define prism , the angle of prism , Refraction Through Prism ,Prism Formula ( Condition for Minimum Deviation Define Prism and the angle of prism :  A simple prism is a homogeneous transparent refracting medium bounded by at least two non parallel plane surfaces inclined at some angle. The surface on which light is incident and another surface from which light comes out should be non parallel.The two non parallel plane surfaces participating in refraction of light are called refracting surfaces and the line of intersection of the two refracting surfaces is called refracting edge. The angle between two refracting surfaces is called the angle of prism or refracting angle and is denoted by A. Refraction Through Prism: Discuss refraction of light due to a prism and hence show that $\delta + A = i + e$, Determination of angle of deviation ($\delta$). Let ABC be the principal section of a prism of refracting angle A. Let a ray of light DE be incident on the refracting surface AB ...

What is a compound microscope ? With the help of a labelled ray diagram, show the image formation by a compound microscope. Derive an expression for its magnifying power. Compound Microscope : A compound microscope consists of two suitable lenses to give large magnification by compounding the magnification given by the two lenses. Construction : compound microscope consists of two convex lenses called objective and the eye piece. An objective lens is of small aperture and small focal length and faces the object to be seen. It forms a real, inverted and magnified image of an object. An eyepiece is a convex lens of large aperture and large focal length as compared to objective. It gives enlarged and virtual image by compounding the effect of the objective. Principle : When an object is placed in front of a convex lens O at a distance between $F_o$ and $2F_o$, the real, inverted and magnified image is formed on the other side of this lens. If this image lies within the focal length of a...

Define Astronomical Telescope , it's construction and principal. Expression for magnifying power ( Angular Magnification) For Distinct Vision Adjustment and For normal Adjustment  Construction: It consists of two achromatic convex lenses mounted co-axially in two metallic tubes. The lens facing the object (which is at infinity) is called objective lens. It has large aperture and large focal length. The other lens through which the image is observed is called eyepiece. It is of small aperture and has small focal length. The tube having eyepiece can be moved in and out of the tube holding objective lens with the help of rack and pinion arrangement. Principle and Theory: Principle of astronomical telescope can be discussed by considering two extreme cases:  (a) When the final image is formed at least distance of distinct vision ( Distinct Vision Adjustment) (b) When final image is formed at infinity (Normal adjustment). Define magnifying power of an astronomical telescope. Deriv...

Derivation - Lens Maker's Formula | Refraction by a lens - Physicskund  Derivation : Consider a lens made of a material of absolute refractive index $n_2$. This lens is placed in a medium of absolute refractive index $n_1$ ($n_1 < n_2$). The lens is bounded by two spherical refracting surfaces $XP_1Y$ and $XP_2Y$. $C_1$ and $C_2$ be their centres of curvature and $R_1$ and $R_2$ be their radii of curvature respectively. $C$ is the optical centre of the lens. STEP 1. Refraction at Surface $XP_1Y$ : Let O be a point object lying in the rarer medium on the principal axis of the refracting surface $XP_1Y$. The incident ray OA after refraction at A bends towards the normal $AC_1$ and meets the principal axis at $I_1$ if the second surface $XP_2Y$ were not present. So, $I_1$ is the real image of the object O. Since object lies in the rarer medium, so we have $-\frac{n_1}{u} + \frac{n_2}{v_1} = \frac{n_2 - n_1}{R_1} \qquad ... (i)$ STEP 2. Refraction at Surface $XP_2Y$ :} In fact, the ...

1. Refraction at a Refracting Surface when Object lies in Rarer Medium :  Question : Prove that $\frac{-n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R}$ When image formed is real :  Consider a convex spherical refracting surface of refractive index $n_2$. Let it be placed in a rarer medium of refractive index $n_1$ ($n_2 > n_1$). A point object O lies in rarer medium on the principal axis at a distance $u$ from the pole of the convex refracting surface. A ray of light from O incident on the convex surface at A. Let C be the centre of curvature, then AC is the normal to the convex surface. After refraction at A, the ray enters the denser medium and bends towards the normal. The refracted ray meets the principal axis at I which is the real image of the object O. The distance of the image I from the pole of the convex surface is $v$. STEP 1. Determination of $i$ and $r$. Let $\alpha$, $\beta$ and $\gamma$ be the angles made by the incident ray, refracted ray and the normal respe...

TOTAL INTERNAL REFLECTION (TIR) OF LIGHT :  Question: What is total internal reflection and critical angle? State the conditions under which total internal reflection of light occurs. Find the relation between critical angle and refractive index of a medium. When a ray of light from point O in the denser medium falls at Q point on the interface separating denser and rarer medium, it is refracted along QQ'. As the angle of incidence increases say at point S, the refracted ray bends towards the interface (say along SS'). At a particular angle of incidence ($\theta_c$), the refracted ray TU travels along the surface of the interface and the angle of refraction is 90$^\circ$ (i.e. r = 90$^\circ$). The angle of incidence corresponding to which angle of refraction becomes 90$^\circ$ is called critical angle ($\theta_c$). When the angle of incidence becomes greater than the critical angle, there is no refraction of light and whole of the incident light is reflected back to the denser ...

Derive a relation between the focal length and radius of curvature stating the assumptions made in the case of a spherical mirror. Assumptions made :  (a) Small aperture approximation:} Aperture of the spherical mirror is assumed to be small. (b) Small angle approximation: Incident ray makes a very small angle with the principal axis and strikes the reflecting surface close to the pole of the mirror. Such a ray is called paraxial ray. (c) Mirror is made of thin refracting material to avoid multiple reflecting. Derivation: Let a ray OA, travelling parallel to the principal axis, incident on a concave mirror at point A (Figure 11). After reflection, the ray passes through the focus (F) of the mirror. CA is the normal to the mirror at A. According to the law of reflection, $i = r = \theta$ Also, $\angle AFP$ is the external angle of  $\triangle ACF$, so  $\angle AFP = \angle ACF + \angle CAF$ $\angle AFP= \theta + \theta = 2\theta$ Now draw AN perpendicular to the principal ...

Speed of image formed by a spherical mirror related to speed of object. Using mirror formula,  $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$,  where $f$ is constant Differentiating both sides w.r.t. 't', we get $\frac{d}{dt}(\frac{1}{u} + \frac{1}{v})=\frac{d}{dt}(\frac{1}{f})$ $-\frac{1}{u^2} \frac{du}{dt} - \frac{1}{v^2} \frac{dv}{dt} = 0$ $ \frac{du}{dt} = -\left(\frac{u}{v}\right)^2 \frac{dv}{dt}$ Here $\frac{dv}{dt} = V_i$ is speed of image and $V_o = \frac{du}{dt}$ is speed of object. $V_o = -\left(\frac{u}{v}\right)^2 V_i$ $V_i = -\left(\frac{v}{u}\right)^2 V_o$ $V_i = -\left(\frac{f}{u-f}\right)^2 V_o$

Mirror Formula and Magnification for Convex Mirror :  mirror formula for a convex mirror. Let AB be an object lying on the principal axis of the convex mirror of small aperture. A'B' is the virtual image of the object lying behind the convex mirror. As $\triangle ABC$ and $\triangle A'B'C$ are similar $\frac{A'B'}{AB} = \frac{CA'}{CA} \quad \cdots (i)$ As $\triangle ABP$ and $\triangle A'B'P$ are similar $\frac{A'B'}{AB} = \frac{PA'}{PA} \quad \cdots (ii)$ From eqns. (i) and (ii), we get, $\frac{CA'}{CA} = \frac{PA'}{PA}$ According to the new Cartesian sign convention all distances are to be measured from pole so $\frac{PC + PA'}{PC - PA} = \frac{PA'}{PA} \quad \cdots (iii)$ Applying sign conventions,  $PC = R$, $PA' = v$; $PA = -u$ Hence, eqn. (iii) becomes, $\frac{R - v}{R - u} = \frac{v}{-u}$ $-uR + uv = vR - vu$ or $vR + uR = 2vu$ Dividing both sides by $uvR$, we get $\frac{1}{v} + \frac{1}{u} = \frac{2}{R}$ But R ...

Question: Derive mirror formula and Magnification for a concave mirror when a real and virtual image is formed by it. 1. mirror formula and Magnification for a concave mirror when a real image is formed by it. As $\triangle ABC$ and $\triangle A'B'C$ are similar, $\frac{A'B'}{AB} = \frac{CA'}{CA} \quad \cdots (i)$ Also $\triangle ABP$ and $\triangle A'B'P$ are similar, $\frac{A'B'}{AB}= \frac{PA'}{PA} \quad \cdots \text(ii)$ From eqns. (i) and (ii), we get $\frac{CA'}{CA}= \frac{PA'}{PA} \quad \cdots (iii)$ According to the new cartesian sign conventions, all the distances are measured from the pole of the mirror. $CA' = (PC - PA')$ and $CA = (PA - PC)\quad \cdots (iv)$ Substituting the values of eqn. (iv) in eqn. (iii), we get, $\frac{(PC - PA')}{(PA - PC)}= \frac{PA'}{PA}$ Applying new cartesian sign convention, $PA' = -v$; $PC = -R$ and $PA = -u$ Hence, eqn. (v) becomes, $\frac{-R + v}{-u + R} = \frac{-v}{-u} \quad ...

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments  9.1 A small candle. 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirroг should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?  Solution:  Here , h = 2.5 cm , u = - 27 cm , R = - 36 cm  Since $f = \frac{R}{2}$ . Therefore, $f = \frac{-36}{2}$ = -18 cm (i) using mirror formula  $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$ , we get $\frac{1}{v}  = \frac{1}{f} - \frac{1}{u}$ $\frac{1}{v} = \frac{1}{(-18)} - \frac{1}{(-27)}$  $\frac{1}{v} = -\frac{1}{18} + \frac{1}{27}$  $\frac{1}{v} = \frac{-3+2}{54}$ $\frac{1}{v} = -\frac{1}{54}$ $v = - 54 cm$ ( Position of image )  Thus , the screen should be placed at a distance of 54 cm in front of the concave mirror....