Given below is a last minute revision note / cheat sheet / ready reckoner of the Geometrical optics section for the IIT JEE Main / Advanced Physics paper. *Students should note that the material is not meant to be a comprehensive review of the section. Only those points which we feel are more relevant in the exam point of view are included* .

- Light experienced by us is the
*effect of neural stimulation of certain cell types*( cones & rods ) caused by a*paticular section ( visible ) of the electromagnetic spectrum*. The point is ,*we as such doesn’t see light , we experience it*… for eg. , rays of light one see when shining a torch in a dark room is due to the scattering of light by dust particles. The inference here is that only when an object scatters / reflects light we will be able to see it . If the light rays pass through an object without any change in it’s nature , i.e if it’s perfectly transparent , we will not be able to see the object. Interestingly , if an objects reflects light perfectly [ say a perfect mirror ] , then again one will be seeing the reflected light which will not give any idea of the said object.*Hence , to see an object , it has to partially reflect or absorb the incident light*. - According to
**Ray optics**, light consists of particles / corpuscles travelling in straight lines . This branch of optics can explain reflection , refraction and dispersion. - According to
**Wave optics**, light is of an electromagnetic wave like nature. It can explain phenomenon like diffraction,interference and polarization. - According to the
**Quantum theory**, light consists of quantized particles called photons . Their momentum and energy are finite and have zero rest mass. Light has the properties of both particles and waves. This theory accounts for phenomena like photoelectric effect and spectral lines. - For real objects , the incident rays diverge from a point [ converges in virtual objects ] . In the case of real images , the rays meet a point after their contact with the optical element , hence the
*image can be seen on a screen*[ for virtual images , the rays appear to come from a point , hence such images cannot be captured on a screen ] . - Eventhough virtual images cannot be captured on a screen ,
*we can see ourselves in front of a plane mirror*–*this is because of the focusing of the divergent rays by the eye lens*. Same is the case with cameras. - A light ray , in a
*homogeneous media*, propagates in a*straight line*and*doesn’t interfere*with another ray at intersections . The ray also can*retrace it path*on reflection . - When light is incident normally , i = r = 0 and δ = 180° . If the incident ray strikes tangentially, i = r = 90° and δ = 0°.
- The angle of deviation of an incident ray getting reflected at an angle i can be calculated by
**δ = π – 2 i**or**δ = π + 2 i**. - When a mirror is rotated through an angle θ , in an axis which lies in the plane of the mirror and perpendicular to the plane of incidence , a fixed incident ray gets reflected through an angle 2 θ.
- An extended object in front of a plane mirror forms an image such that a) size of the image and object are equal b) parallely placed object creates an upright image and c) perpendicularly placed object creates an inverted image .
- For an observer to see his full image on a plane mirror , the mirror has to be at leat half the size of the observer.
- For a plane mirror , velocity of an object is equal to the velocity of the related image , when parallel to the mirror surface. If
**v**_{o m}= velocity of object with respect to mirror ,**v**_{i m}= velocity of image w.r.t mirror ,**v**_{i}= velocity of image w.r.t ground and**v**_{o}= velocity of object w.r.t ground , then ,**v**_{o m}**= v**_{o}**– v**_{m}and**v**_{i m}**= v**_{i}**– v**_{m}. - To find the number of images formed by two inclined plane mirrors , inclined at an angle θ , a) if 360°/θ = even number , then the number of images = 360°/θ – 1 , and b) if 360°/θ = odd number , then the no of images = 360°/θ – 1 , if the object is placed on the angle bisector and it’s 360°/θ if the object is not placed on the angle bisector .
- The
**sign convention**used to solve problems in the case of spherical mirrors , for both reflection and refraction are a) distances are to be measured from the pole b) distances measured in the direction of incident rays are taken as positive c) distances measured opposite to the direction of incident rays are taken as negative and d) distances above the principal axis are taken as positive and those below the principal axis are taken as negative. - To find the image of an object graphically ,
– a) the ray passing through the centre of curvature which gets reflected back along itself b) the ray initially parallel to the principal axis which gets reflected through the focus of the mirror c) the ray initially passing through the focus which gets reflected parallel to the principal axis and d) the ray incident at the pole which is reflected symmetrically.**any two**of the following rays are to be considered - The optical power of a mirror is
**P = – 1/f**where f is the focal length in metres. Note that P is expressed in diopters. - If fo – focal length of a mirror and x is the distance of the object from the mirror , then , magnification for a concave mirror is
**m = fo/( fo – x )**and that of a convex mirror is**m = fo/( fo + x )**. Note – if a virtual object is placed between F and P , then a*convex mirror gives a real image*. Here ,**m = – f/( fo – x ).** - For spherical mirrors , the object and image velocities are related by
**v**_{i}**= – m² v**_{o}where**v**_{i}= velocity of the image w.r.t the mirror ,**v**_{o}= velocity of the object w.r.t the mirror and m = lateral / transverse magnification**a)**if the object is at the centre of curvature ,**|dv/dt| = |-du/dt**| ,**b)**if the object is moving between the focus and the centre of curvature**|dv/dt| > |-du/dt|****c)**if the object is moving between the focus and the mirror pole , then**[dv/dt ] = [ v²/u² ] [du/dt ]**here , as |m| > 1 , the speed of the image will be greater than the speed of the object and**d)**for a small object placed along the principal axis ,**dv = – m² du**. [ Note – in all the above cases , |dv/dt| and |du/dt| are velocities w.r.t the mirror and not with that of the ground ]. - Differentiation of the mirror formula gives
**dv/du = v²u²**. Here dv/du is called longitudinal magnification .*It should be noted that in this case , the focus should not lie between the initial and final points of the object.* - For an object moving perpendicular to the principal axis , h2/h1 = v/u , hence
**h2 = ( -v/u) h1**. Here , if we are dealing with a point object , then as the x co-ordinates of both the object and image are constant , we get on differentiation ,**dh2/dt = – [ v/u ] [ dh2/dt]** - Newton’s formula is useful when distances are asked related to the focus .
**XY = f ²**, where X and Y are distances along the principal axis of the object and the image respectively. - When using Snell’s law in cases of refraction , use
**n**_{1}**sin i = n**_{2}**sin r .**Refraction doesn’t change the frequency of the light. The refractive index of a medium relative to vaccum is**√ (μ**_{r}**ε**_{r}**)** - Apparent shift of an object due to refraction is given by = real depth
**x [ 1 – μ**_{2}**/μ**_{1}**]**. If μ1 < μ2 , then image distance is greater than object distance and the shift is negative. - If the angle of inidence is small ,the apparent shift of an object due to refraction is
**d’ = d / n**, here d’ = apparent depth , d = real depth and n = ni/nr = refr.index of incident medium / refr.index of refractive medium. - Apparent shift of an object due to refraction of a parallel slab is given by
**s = [ x+ t/μ ] – [ (x+t)] = t [ 1 – ( t/μ ) ]**. The object appears to shift along the perpendicular of the slab by a distance t [ 1 – ( t/μ ) ] .Note that the final image is at a distance**x + ( 1/μ )**behind the second interface.