Spherical Mirror: Highly polished curved surface whose reflecting surface is cut from part of a hollow sphere is called spherical mirror.

Concave mirror: The spherical mirror whose reflecting surface is curved inwards towards the center of the sphere.

Convex mirror: The spherical mirror whose reflecting surface is curved outwards away from the center of the sphere.

Pole: Mid point of the spherical mirror.Radius of curvature: It is radius of hollow sphere of which spherical mirror is formed.Aperture: It is part of mirror exposed to incident light or diameter of reflecting surface of spherical mirror.Principal Axis: It is a line joining pole and center of curvature of sphetical mirror.Principal Focus: It is the point on principal axis of spherical mirror where all the parallel ray of lights parallel to principal axis meet or appear to meet. Focus Plane: It a plane that is normal and perpendicular to principal axis and passes thru prinicipal focus.Focul Length: It is distance between pole and the prinicipal focus of a spherical mirror.

Linear magnification, also called lateral magnification or transverse magnification.

Cartesian sign convention for measuring distance in spherical mirror:1. All distances are measured from the pole of the mirror or optical center of lens.2. The distances measured in the same direction of incident light are taken as positive and those measured in the direction of opposite to the direction of incident light are taken as negative.3. The height measured upward with respect to x-axis and normal to the pricipal axis of mirror or lens are taken as positive. The height measured downwards are taken negative.4. Object are always placed at left side of mirror.

Refraction:When a beam of light encounters another transparent medium, a part of light gets reflected back into the first medium while the rest enters the other. The direction of propagation of an obliquely incident ray of light that enters the other medium, changes. This phenomenon is called refraction.

Snell's law of refraction(i) The incident ray, the refracted ray and the normal to the interface at the point ofincidence, all lie in the same plane.(ii) The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant.n21=refractive index of medium 2 with respect to 1=sini

sinr=speed of light in medium 1

speed of light in medium 2$n}_{21}=refractive\phantom{\rule{0.22em}{0ex}}index\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}medium\phantom{\rule{0.22em}{0ex}}2\phantom{\rule{0.22em}{0ex}}with\phantom{\rule{0.22em}{0ex}}respect\phantom{\rule{0.22em}{0ex}}to\phantom{\rule{0.22em}{0ex}}1=\frac{sini}{sinr}=\frac{speed\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}light\phantom{\rule{0.22em}{0ex}}in\phantom{\rule{0.22em}{0ex}}medium\phantom{\rule{0.22em}{0ex}}1}{speed\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}light\phantom{\rule{0.22em}{0ex}}in\phantom{\rule{0.22em}{0ex}}medium\phantom{\rule{0.22em}{0ex}}2$n21=1

Optical Density: It is the ratio of the speed of light in two medium

Total Internal Reflection:When 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.If the angle of incidence is increased still further, refraction is not possible, and the incident ray is totally reflected. This is called total internal reflection.

Critical angle: The critical angle is the angle of incidence where the angle of refraction is 90°. The light must travel from an optically denser medium to an optically rarer medium.

Total Internal Reflection and its application1. Mirage2. Diamond3. Prism4. Optical Fibres

Cauchy's Formula ( relationship between the refractive index and wavelength of light)𝜇=A+B

Effective Power if several thin lenses are in contactP=P1+P2+P3+....$P={P}_{1}+{P}_{2}+{P}_{3}+.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.44em}{0ex}}$

Total magnification (m)m=m1. m2. m3. m4.......$m={m}_{1}\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}{m}_{2}\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}{m}_{3}\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}{m}_{4}\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.44em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}$

An aberration is a distortion in an image.(i) Spherical aberration is an optical problem. It occurs when all incoming light rays focus at different points after passing thru lens.Spherical aberration can be minimized by use of aspherical or non-spherical lens surface. This type of lens curves outwards on one side so that it converges the light rays into a single focal point.Spherical aberration can be reduced by usinga. lens of large focal lengthsb. plano-convex lensesd. crossed lensese. combining convex and concave lensf. use of aspherical or non-spherical lens surface. (ii) Chromatic Aberration Image of a white object formed by lens is usually coloured and blurred. This defect of the image produced by lens is called chromatic aberration.Refraction thru Prism:The angle between the emergent ray and the direction of incident ray is called angle of deviation.Angle of deviation=𝛿=i+e-A $=\mathit{\delta}=i+e-A\phantom{\rule{0.22em}{0ex}}$where i=angle of incidencee=angle of emergenceA=vertex angle of prism (refracting angle of prism) When i=e then angle of deviation is minimum.awhen , 𝛿=Dmthen i=e , r1=r2=rangle of refraction = rr=A

2i=A+Dm

2n21=sin i

sin r=sinA+Dm

2

sinA

2$\begin{array}{l}when\phantom{\rule{0.22em}{0ex}},\phantom{\rule{0.22em}{0ex}}\mathit{\delta}={D}_{m}\\ then\phantom{\rule{0.44em}{0ex}}i=e\phantom{\rule{0.22em}{0ex}},\phantom{\rule{0.44em}{0ex}}{r}_{1}={r}_{2}=r\\ angle\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}refraction\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}r\\ r=\frac{A}{2}\\ \\ i=\frac{A+{D}_{m}}{2}\phantom{\rule{0.22em}{0ex}}\\ {n}_{21}=\frac{sin\phantom{\rule{0.22em}{0ex}}i}{sin\phantom{\rule{0.22em}{0ex}}r}=\frac{sin\frac{A+{D}_{m}}{2}}{sin\frac{A}{2}}\\ \end{array}$Deviation produced by thin prism𝛿m=(n-1)A${\mathit{\delta}}_{m}=(n-1)A$Dispersion of light: The phenomenon of splitting of light into its component colours is known asdispersion.The pattern of colour components of light is called the spectrum of light. Angular dispersion: The angle between the emergent rays of any two colours is called angular dispersion between those rays.Dispersive power : When a white light passes thru a thin prism, the ratio of the angular dispersion between the violet and the red emergent rays and the deviation suffered by a mean ray (ray of yellow colour) is called the "dispersive power" of the material of the prism.Dispersive power=𝜔=nV-nC

nY-1$Dispersive\phantom{\rule{0.22em}{0ex}}power=\mathit{\omega}=\frac{{n}_{V}-{n}_{C}}{{n}_{Y}-1}$Rainbow:The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This is a phenomenon due to combined effect of dispersion, refraction and reflection of sunlight by spherical water droplets of rain. Primary RainbowThe primary rainbow is formed when there is refraction and a single reflection inside the water droplet.Sunlight is first refracted as it enters a raindrop, which causes the different wavelengths (colours) of white light to separate. Longer wangelength of light (red) are bent the least while the shorter wavelength (violet) are bent the most.Then these component rays strike the inner surface of the water drop and get internally reflected if the angle between the refracted ray and normal to the drop surface is greater then the critical angle (48 degree). The reflected light is refracted again as it comes out of the drop.So the primary rainbow is a result of three-step process, that is, refraction, reflection and refraction.Secondary RainbowWhen light rays undergoes two internal reflections inside a raindrop, instead of one as in the primary rainbow, a secondary rainbow is formed. It is due to four-step process. The intensity of light is reduced at the second reflection and hence the secondary rainbow is fainter than the primary rainbow. The order of the colours is reversed in secondary rainbow.Between primary and secondary rainbow there is dark band known as Alexander dark band. It happens becausea. Light is scattered into this region interfere destructively.b. The angle made at the eye by the scattered ray with respect to the incident light of the sun lies between 42 degree to 50 degree.Microscope1. Simple Microscope2. Compound MicroscopeObjective: The lens nearer to the object is called objectiveEyepiece: It functions like a simple microscope or magnifier which produces final enlarged and virtual image.Simplle MicroscopeA simple magnifier or microscope is a converging lens of small focal length.aMagnification=m=v

u=v(1

v-1

f)=(1-v

f)v=-D=-25 cm$\begin{array}{l}Magnification=m=\frac{v}{u}=v(\frac{1}{v}-\frac{1}{f})=(1-\frac{v}{f})\\ \\ v=-D=-25\phantom{\rule{0.22em}{0ex}}cm\end{array}$Magnification of simple magnifier (simple Microscope)= (1+D

f)$(1+\frac{D}{f})\phantom{\rule{0.22em}{0ex}}$Angular magnification is the ratio of the angle subtended by the image to that subtended by the object, if object is placed at D (D=25 cm)aangular magnification=𝜃image

𝜃objectangular magnification=height of image

distance of image from eyw

height of object

D$\begin{array}{l}\\ angular\phantom{\rule{0.22em}{0ex}}magnification=\frac{{\mathit{\theta}}_{image}}{{\mathit{\theta}}_{object}}\phantom{\rule{0.22em}{0ex}}\\ \\ angular\phantom{\rule{0.22em}{0ex}}magnification=\frac{\frac{height\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}image}{distance\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}image\phantom{\rule{0.22em}{0ex}}from\phantom{\rule{0.22em}{0ex}}eyw}}{\frac{height\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}object}{D}}\phantom{\rule{0.22em}{0ex}}\end{array}$Angular Magnification when the image is at infinityaLinear Magnification=m=himage

hobject= v

u tan𝜃i=himage

-v=hobject. v

-v , u =hobject

-u≈𝜃 when image is at infinity then u=-f so, tan𝜃i=hobject

f⟹𝜃i=hobject

f𝜃o=hobject

Dangular magnification when image is at infinity=𝜃i

𝜃o= D

f $\begin{array}{l}Linear\phantom{\rule{0.22em}{0ex}}Magnification=m=\frac{{h}_{image}}{{h}_{object}}=\frac{\phantom{\rule{0.22em}{0ex}}v\phantom{\rule{0.22em}{0ex}}}{\phantom{\rule{0.22em}{0ex}}u\phantom{\rule{0.22em}{0ex}}}\phantom{\rule{0.44em}{0ex}}\\ tan{\mathit{\theta}}_{i}=\frac{{h}_{image}}{-v}=\frac{{h}_{object}\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}v}{-v\phantom{\rule{0.22em}{0ex}},\phantom{\rule{0.22em}{0ex}}u\phantom{\rule{0.22em}{0ex}}}\phantom{\rule{0.22em}{0ex}}=\frac{{h}_{object}}{-u}\approx \mathit{\theta}\phantom{\rule{0.22em}{0ex}}\\ \\ when\phantom{\rule{0.22em}{0ex}}image\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}at\phantom{\rule{0.22em}{0ex}}infinity\phantom{\rule{0.22em}{0ex}}then\phantom{\rule{0.22em}{0ex}}u=-f\phantom{\rule{0.22em}{0ex}}\\ so,\phantom{\rule{0.44em}{0ex}}tan{\mathit{\theta}}_{i}=\frac{{h}_{object}}{f}\phantom{\rule{0.22em}{0ex}}\\ \u27f9{\mathit{\theta}}_{i}=\frac{{h}_{object}}{f}\phantom{\rule{0.22em}{0ex}}\\ {\mathit{\theta}}_{o}=\frac{{h}_{object}}{D}\phantom{\rule{0.22em}{0ex}}\\ \\ angular\phantom{\rule{0.22em}{0ex}}magnification\phantom{\rule{0.22em}{0ex}}when\phantom{\rule{0.22em}{0ex}}image\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}at\phantom{\rule{0.22em}{0ex}}infinity=\frac{{\mathit{\theta}}_{i}}{{\mathit{\theta}}_{o}}\phantom{\rule{0.22em}{0ex}}=\frac{\phantom{\rule{0.22em}{0ex}}D\phantom{\rule{0.22em}{0ex}}}{\phantom{\rule{0.22em}{0ex}}f\phantom{\rule{0.22em}{0ex}}}\\ \end{array}$Compound Microscope1. The lens nearest to the object is called objective. It forms real inverted magnified image of the object. It serves as object for the eyepiece.2. The final image formed by the eyepiece is large and virtual.3. Magnification produced by objective=mo=v0

u0$={m}_{o}=\frac{{v}_{0}}{{u}_{0}}\phantom{\rule{0.22em}{0ex}}$4. Magnification produced by eyepiece=me=(1+D

fe)$={m}_{e}=\phantom{\rule{0.22em}{0ex}}(1+\frac{D}{{f}_{e}})\phantom{\rule{0.44em}{0ex}}$5. Magnification of microscope=m=m0me=vo

u0(1+D

fe)$=m={m}_{0}{m}_{e}=\frac{{v}_{o}}{{u}_{0}}(1+\frac{D}{{f}_{e}})\phantom{\rule{0.22em}{0ex}}$L = tubelength of microscope = It is the distance between the image-side focal point of the objective and the object-side focal point of the eyepiece.Magnification in terms of tubelengtham0=h'

h=L

f0me=(1+D

fe)$\begin{array}{l}{m}_{0}=\frac{h\text{'}}{h}=\frac{L}{{f}_{0}}\phantom{\rule{0.44em}{0ex}}\\ \\ {m}_{e}=(1+\frac{D}{{f}_{e}})\phantom{\rule{0.22em}{0ex}}\end{array}$If we consider final image to be formed at near point of the eye then for eyepiece. ame= D

fem=m0me=L

f0.(1+D

fe)$\begin{array}{l}{m}_{e}=\frac{\phantom{\rule{0.22em}{0ex}}D\phantom{\rule{0.22em}{0ex}}}{\phantom{\rule{0.22em}{0ex}}{f}_{e}\phantom{\rule{0.22em}{0ex}}}\phantom{\rule{0.22em}{0ex}}\\ m={m}_{0}{m}_{e}=\frac{L}{{f}_{0}}\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}(1+\frac{D}{{f}_{e}})\phantom{\rule{1.1em}{0ex}}\end{array}$If we consider final image is formed at infinity so we have to use angular magnification for eyepiece. ame= D

fem=m0me=L

f0D

fe$\begin{array}{l}{m}_{e}=\frac{\phantom{\rule{0.22em}{0ex}}D\phantom{\rule{0.22em}{0ex}}}{\phantom{\rule{0.22em}{0ex}}{f}_{e}\phantom{\rule{0.22em}{0ex}}}\phantom{\rule{0.22em}{0ex}}\\ m={m}_{0}{m}_{e}=\frac{L}{{f}_{0}}\frac{D}{{f}_{e}}\phantom{\rule{0.88em}{0ex}}\end{array}$TelescopeA telescope is used to provide angular magnification of of distinct objectsRefracting telescopeIt has eyepiece and objective.Objective has large focus length and larger aperture than eyepiece.1. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. 2. The eyepiece magnifies this image producing a final inverted image at infinity.3. 𝛼$\mathit{\alpha}$ = angle subtended at the lens by the object.atan𝛼=h

fo⟹𝛼≈h

fo$\begin{array}{l}tan\mathit{\alpha}=\frac{h}{{f}_{o}}\phantom{\rule{0.22em}{0ex}}\\ \u27f9\mathit{\alpha}\approx \frac{h}{{f}_{o}}\end{array}$4. 𝛽$\mathit{\beta}$ = angle subtended at the eyepiece lens by the image.atan𝛽=h

fe⟹𝛽≈h

fe$\begin{array}{l}tan\mathit{\beta}=\frac{h}{{f}_{e}}\phantom{\rule{0.22em}{0ex}}\\ \u27f9\mathit{\beta}\approx \frac{h}{{f}_{e}}\end{array}$5. The magnifying power m is the ratio of the angle 𝛽$\mathit{\beta}$ subtended at the eye by the final image to the angle 𝛼$\mathit{\alpha}$ which the object subtends at the lens or the eye. Hencem=𝛽

𝛼=h

fefo

h=fo

fe$m=\frac{\mathit{\beta}}{\mathit{\alpha}}=\frac{h}{{f}_{e}}\frac{{f}_{o}}{h}=\frac{{f}_{o}}{{f}_{e}}\phantom{\rule{0.22em}{0ex}}$6. Length of the telescope=fo+fe$={f}_{o}+{f}_{e}\phantom{\rule{0.22em}{0ex}}$Reflecting Telescope (Cassegrain telescope)1. Telescopes with mirror objectives are called reflecting Telescope.Advantages of reflecting telescopea. There is no chromatic aberation in mirror.b. If parabolic reflecting surface is chosen then spherical abheration is also removed.c. Reflectors are comparatively very light in weight than the lens. This allows us to create very large size reflectors which is not possible in case of lens.d. The image is brighter compared to that in a refracting type telescope.e. High resolution is achieved by using a mirror of large aperture.f. Mirror requires grinding and polishing of only one side.Magnifying power=m=fo

fe$Magnifying\phantom{\rule{0.22em}{0ex}}power=m=\frac{{f}_{o}}{{f}_{e}}\phantom{\rule{0.22em}{0ex}}$Human EyeHuman Eye: Human eye is an optical instrument which forms real image of the objects on retina.Cornea: It is a curved from surface through which light enters the eye.Pupil: A small opening in the iris is known as a pupil. Pupil controls the amount of light that enters the eye.Retina: It is a light-sensitive layer that consists of numerous nerve cells. It converts images formed by the lens into electrical signals that are transmitted to the brain through optic nerves.1. The closest distance for which the lens can focus light on the retina is called the least distance of distinct vision, or the near point.2. Least distance of distinct vision is the minimum object distance that is able to produce a distinct image on the retina. It is denoted by D, which is 25 cm for normal vision.Myopia: Near-sightedness : 1. A person with Myopia can see nearby objects clearly2. A person with myopia cannot see faraway objects clearly.3. The far point for the myopic eye is nearer than infinity4. It occurs due to excessive curvature of the eye lens elongation of eyeball5. The image of a distance object is formed in front of the retina and not on the retina6. This defected is corrected by using Concave lenses such that the lens will bring the image back on to the retina.Hypermetropia : Far sightedness : 1. A person with Hypermetropia can see far away objects clearly.2. A person with Hypermetropia cannot see nearby objects clearly.3. The near point of the eye is more than 25 cm4. It happens mostly during latter stages in life, as a result of the weakening of the ciliary muscles and/or the decreased flexibility of the lens.5. It happens when focal length of the eye lens is too long6. It happens when eyeball has become too small. 7. The image of a distance object is formed in behind the retina and not on the retina8. This defected is corrected by using Convex lenses such that the lens will bring the image back on to the retina.Presbyopia1. The power of accommodation of the eye usually decreases with ageing. The ciliary muscles weaken and thereby the flexibility of the eye lens reduces.2. The near point moves away.3. It is corrected by converging lens. So convex lenses are recommendedMyopia and Hypermetropia1. Sometimes a person may suffer from both near sightedness and far-sightedness.2. This is corrected by using bifocal lenses.3. Bifocal lenses consists of concave on the upper portion and convex on the lower portion.4. Concave supports distinct vision and convex supports near visionScatteringScattering of light is the phenomenon in which light rays get deviated from its straight path on striking an obstacle like dust or gas molecules, water vapours.Light of shorter wave length is scattered more than light of longer wavelength.Rayleigh scattering: Lights of shorter wave lengths is scattered much more than light of longer wavelengths.The amount of scattering is inversely proportional to the fourth power of the wavelength. This is known as Rayleigh scattering.Intensity of the scattered light is inversely propertional to the fourth power of the wavelength, provided the scatterer is smaller than the wavelength of light.Scattering∝1

𝜆4$Scattering\propto \frac{1}{{\mathit{\lambda}}^{4}}$Phenomenon based upon scattering of light1. Sky appears blueAs sunlight travels through the earth’s atmosphere, it gets scattered (changes its direction) by the atmospheric particles. Light of shorter wavelengths is scattered much more than light of longer wavelengths. (The amount of scattering is inversely proportional to the fourth power of the wavelength.). Hence, the bluish colour predominates in a clear sky, since blue has a shorter wave-length than red and is scattered much more strongly. In fact, violet gets scattered even more than blue, having a shorter wavelength. But since our eyes are more sensitive to blue than violet, we see the sky blue.2. Clouds appear whiteLarge particles like dust and water droplets present in the atmosphere behave differently. The relevant quantity here is the relative size of the wavelength of light 𝜆$\mathit{\lambda}$, and the scatterer (of typical size, say, a). For a≪𝜆$a\ll \mathit{\lambda}$ one has Rayleigh scattering which is proportional to 1

𝜆4$\frac{1}{{\mathit{\lambda}}^{4}}$For a≫𝜆$a\gg \mathit{\lambda}$ , i.e., large scattering objects (for example, raindrops, large dust or ice particles) this (Rayleigh scattering) is not true. In this case all wavelengths are scattered nearly equally. Thus, clouds which have droplets of water with a≫𝜆$a\gg \mathit{\lambda}$ are generally white.Water vapours are much larger than air molecules and all colors of light are scattered equally. Thus all colors of light combine to make white light.3. Reddish appearance of the sun at sunrise and sunsetAt sunset or sunrise, the sun’s rays have to pass through a larger distance in the atmosphere. Most of the blue and other shorter wavelengths are removed by scattering. The least scattered light reaching our eyes, therefore, the sun looks reddish. This explains the reddish appearance of the sun and full moon near the horizon.4. Red light is used in danger signalIn the visible spectrum, the red color has the largest wavelength. The red colour is least scattered by fog or dust particles. So we can observe red colour easily even in foggy conditions. That is red light is used in danger signal.5. Infra red photography is possible in fog and mistInfrared are used in photography at night and also in mist and fog because they have long wavelength. As they are having low frequency the energy associated with them is also low so they do not scatter much and can penetrate appreciably through it.QUESTIONSQ. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?Answer:aLens type=convex1

v-1

u=1

flet lens is place at distance x from first wall so distance of lens from second wall = 30-xv=+(3-x)u=-x1

3-x+1

x=1

f⟹3f=30x-x2⟹3df

dx=3-2x for focal length to be maximum,df

dx=0⟹3-2x=0⟹x=1.5so, 3f=3⨯1.5-1.5⨯1.5=2.25⟹f=0.75m$\begin{array}{l}Lens\phantom{\rule{0.22em}{0ex}}type=convex\\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ let\phantom{\rule{0.22em}{0ex}}lens\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}place\phantom{\rule{0.22em}{0ex}}at\phantom{\rule{0.22em}{0ex}}distance\phantom{\rule{0.22em}{0ex}}x\phantom{\rule{0.22em}{0ex}}from\phantom{\rule{0.22em}{0ex}}first\phantom{\rule{0.22em}{0ex}}wall\phantom{\rule{0.22em}{0ex}}\\ so\phantom{\rule{0.22em}{0ex}}distance\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}lens\phantom{\rule{0.22em}{0ex}}from\phantom{\rule{0.22em}{0ex}}second\phantom{\rule{0.22em}{0ex}}wall\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}30-x\\ v=+(3-x)\\ u=-x\\ \frac{1}{3-x}+\frac{1}{x}=\frac{1}{f}\phantom{\rule{0.22em}{0ex}}\\ \u27f93f=30x-{x}^{2}\\ \u27f93\frac{df}{dx}=3-2x\phantom{\rule{0.22em}{0ex}}\\ for\phantom{\rule{0.22em}{0ex}}focal\phantom{\rule{0.22em}{0ex}}length\phantom{\rule{0.22em}{0ex}}to\phantom{\rule{0.22em}{0ex}}be\phantom{\rule{0.22em}{0ex}}maximum,\phantom{\rule{0.22em}{0ex}}\frac{df}{dx}=0\\ \u27f93-2x=0\\ \u27f9x=1.5\\ so,\phantom{\rule{0.22em}{0ex}}3f=3\u2a2f1.5-1.5\u2a2f1.5=2.25\\ \u27f9f=0.75m\end{array}$Q.9.7Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?A.9.7Lens Maker formulaa1

f=(n-1)(1

R1-1

R2)Let the radius of curvature is=Rso R1=RR2=-R1

20=(1.55-1)(1

R-1

-R)=(0.55)(2

R)⟹R=0.55⨯2⨯20=22 cm$\begin{array}{l}\frac{1}{f}=(n-1)(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}})\phantom{\rule{0.66em}{0ex}}\\ Let\phantom{\rule{0.22em}{0ex}}the\phantom{\rule{0.22em}{0ex}}radius\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}curvature\phantom{\rule{0.22em}{0ex}}is=R\\ so\phantom{\rule{0.22em}{0ex}}\\ {R}_{1}=R\\ {R}_{2}=-R\\ \frac{1}{20}=(1.55-1)(\frac{1}{R}-\frac{1}{-{R}_{}})\phantom{\rule{0.44em}{0ex}}=(0.55)\left(\frac{2}{R}\right)\\ \u27f9R=0.55\u2a2f2\u2a2f20=22\phantom{\rule{0.22em}{0ex}}cm\\ \end{array}$Q.9.12A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.A.9.12aNote: Here sharp focus is needed.Givenf0=8.0 mm=0.8 cmfe=2.5 cmD=25 cmObjective lensu0=0.9 cm1

v0-1

u0=1

f0⟹1

v0-1

-0.9=1

0.8⟹v0=72

10=7.2 cmEye piece1

ve-1

ue=1

fefe=+2.5 cmve=D=-25 cm1

-25-1

ue=1

2.5⟹ue=-2.27 cmSeparation between two lenses=|ve|+|ue|=7.2+2.27=9.47 cm Magnification=m=v0

Q: Will the focus length of a lens for red light be more , same or less than that for blue light ?A: Focal length of lens for red light will be more the focal length of lens for blue light or violet light.1

f=(n-1)(1

R1-1

R2)$\frac{1}{f}=(n-1)(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}})$Refractive index of blue light is more than red light so focus length of blue light will be less than red light.

Q-An unsymmetrical double convex thin lens, forms the image of a point object on its axis will the position of the image change in the lens is reversed ?A- Lens Maker formula1

f=(n-1)(1

R1-1

R2)$\frac{1}{f}=(n-1)(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}})$By the less, the value of and this signs R1$R}_{1$ and R2$R}_{2$ are reversed.1

f=-(n-1)(1

R2-1

R1)$\frac{1}{f}=-(n-1)(\frac{1}{{R}_{2}}-\frac{1}{{R}_{1}})$So for a given u, v remain unaffected. So position of the image will not change.

Q-A small telescope has an objective lens of focal length 140 cm and eyepiece of focal length 5.0 cm What is the magnifying power of telescope for viewing distant objects when the telescope is in normal adjustment (ie When the final image is at infinity ) What is the seperation between the objective lens and eyepiece ?A-In normal adjustment separation separation between the objective and the eyepiece=fo+fe=140+5=145 cm$={f}_{o}+{f}_{e}=140+5=145\phantom{\rule{0.22em}{0ex}}cm$

QDerive mirror equation for a convex mirror. Using it, show that a convex mirror always produces a virtual image, independent of the location of object.A𝛥A'B'F and 𝛥MPF are similar $\mathit{\Delta}A\text{'}B\text{'}F\phantom{\rule{0.44em}{0ex}}and\phantom{\rule{0.44em}{0ex}}\mathit{\Delta}MPF\phantom{\rule{0.22em}{0ex}}are\phantom{\rule{0.22em}{0ex}}similar\phantom{\rule{0.44em}{0ex}}$B'A'

PM=B'F

FP----------(a)$\frac{B\text{'}A\text{'}}{PM}=\frac{B\text{'}F}{FP}\phantom{\rule{0.44em}{0ex}}----------(a)\phantom{\rule{0.22em}{0ex}}$ PM=AB -----------(b)$\phantom{\rule{0.22em}{0ex}}PM=AB\phantom{\rule{0.22em}{0ex}}-----------\phantom{\rule{0.22em}{0ex}}(b)\phantom{\rule{0.22em}{0ex}}$using (a) and (b) B'A'

AB=B'F

FP$\frac{B\text{'}A\text{'}}{AB}=\frac{B\text{'}F}{FP}\phantom{\rule{0.22em}{0ex}}$a𝛥A'B'P and 𝛥ABP are similar , because∠ABC=∠A'B'P∠APB=∠A'PB' (by lay of reflection)$\begin{array}{l}\mathit{\Delta}A\text{'}B\text{'}P\phantom{\rule{0.44em}{0ex}}and\phantom{\rule{0.44em}{0ex}}\mathit{\Delta}ABP\phantom{\rule{0.22em}{0ex}}are\phantom{\rule{0.22em}{0ex}}similar\phantom{\rule{0.44em}{0ex}},\phantom{\rule{0.22em}{0ex}}because\\ \angle ABC=\angle A\text{'}B\text{'}P\\ \angle APB=\angle A\text{'}PB\text{'}\phantom{\rule{0.44em}{0ex}}(by\phantom{\rule{0.22em}{0ex}}lay\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}reflection)\end{array}$aSo,B'A'

AB=B'P

BP PM=AB -----------(c)$\begin{array}{l}So,\\ \frac{B\text{'}A\text{'}}{AB}=\frac{B\text{'}P}{BP}\phantom{\rule{0.44em}{0ex}}PM=AB\phantom{\rule{0.22em}{0ex}}-----------\phantom{\rule{0.22em}{0ex}}(c)\\ \phantom{\rule{0.22em}{0ex}}\end{array}$By (b) and (c)B'F