Home | Courses | Back to Chapter | Any Question? |

CLASS-11 PHYSICS

Reflection through the plane and Curved Surface.

1. Glancing angle (g): The angle between incident ray and reflecting surface is called glancing angle.
In figure: < AOX is the glancing angle.



2. Deviation angle (δ): The angle between the reflected ray of light and original path of light produced back is called
Deviation angle.
In figure: < BOC is the deviation angle.



3. Relation between glancing angle and deviation angle (g and δ):


< BOC=δ
δ = < BOY + < YOC
δ = < NOY - < NOB + < YOC
δ = 900 -r + g [< XOA=< YOC Vertically Opposite Angle]
But,
< AON=< NOB [From law of reflection]
i=r
900-i+g
[90-i+g]
g+g
2g


Law of rotation of light:



When mirror is rotated by an angle keeping the incident ray fixed then reflected ray of light rotates by an angle
2θ. This is called law of rotation of light.
Let us consider XY is a plane mirror. Let, AO is the incident ray which reflects along OB.
Let, mirror is rotated by an angle ; then incident ray reflect along OB' as shown in figure.
Let, be the glancing angle.
Then from figure,
< BOC=δ1= 2g
< B'OC = δ2= 2(g+θ)
Again from figure,
β= < B'OB = < B'OC - < BOC
β= 2(g+θ)-2g
β= 2g+2θ-2g
β= 2θ
Types of image:
1. real image: inverted, virtual object.
2. virutal image: erect, real object

1. Real image: The image which is obtained by the actual intersection of reflected ray of light or the
refracted ray of light is called real image. It can be produced on screen and it is always inverted.
The real image can be obtained by the plane mirror when the object is virtual which is shown in figure.




2. Virtual image: An image which cannot be obtained on a screen is called virtual image. It is formed
when the light rays after reflection do not actually intersect but intersect when produced backward.
The light rays appear as coming from an image. The virtual image is erect or upright with respect to the object.



Image formed by Plane Mirror:



The image formed by plane mirror is same in size i.e. image distance and object distance are equal.
Let us consider, XY be the plane mirror. Let O be the object. Let OA is the incident ray of light incident normally which reflect along AO.
Let OB be the another incident ray which reflect along BC. Now, let us draw a normal at point B as shown in figure.
Again, let us draw AO and BC back to meet at point I which is virtual image.

Now,
1. < OBn=< nBC [From law of reflection]
2. < AOB=< OBn [alternate angle]
3. < AIB=< nBC [Corresponding angle]
4. < AOB=< AIB [From statement 1 and 2] and < OAB=< IAB [Right angle]
5. AB = AB [common side]
OAB and AIB are congruent. So, OA=AI. This means object distance is equal to image distance.

Relation Between Focal length and Radius of Curvature:



Let us consider a concave mirror of small apperature with focal length (F). Let OA is the incident ray of light incident parallel to
the principal axis at point A which reflects passing through the focus as shown in figure. Let us draw a normal passing through the centre
of curvature at point A.

< OAC=< ACF [alternate angle]
< OAC=< CAF [law of reflection]
< ACF=< FAC
CF=FA [CAF= isosceles triangle]
CP= R= CF+FP = CF+f
FA+f
Since mirror is small in apperature A lies very closed to P. So, we can write,
PF FA


R= F+F = 2F
This shows that focal length of mirror depend only with a radius of curvature.

Mirror formula for concave mirror when image form is real.

Unable to Flash Image

(A'B')/LN = (A'F)/FN
(A'B')/AB = (A'F)/FN <------ (i)

(A'B')/AB = (A'C)/AC <------ (ii)

From equation (i) and equation (ii)
(A'F)/FN = (A'C)/AC <-------- (iii)

PA'= v , PC= R
PA= u , PF= f
Again from figure,
A'C= PC-PA'
AC= PA-PC
A'F= PA'-PF
Now equation (iii) becomes,
[(PC-PA')/(PA-PC) = (PA'-PF)/F]
Taking sign convention,
(R-v)/(u-R) = (v-f)/f
or, (2f-v)/(u-2f) = (v-f)/f [since, R=2f]
2f2-vf = uv-2vf-uf+2f2
or, vf = 2vf + uf - uv
or, vf = uv - uf
or, (vf/uvf) = (uv/uvf) - (uf/uvf)
1/u = 1/f - 1/v
therefore, 1/f = 1/u + 1/v

Mirror formula for convave mirror when image formed is virtual

Unable to flash Image

AP= u= object distance
A'P= -v= image distance (image is virtual)
A'B'= h1= height of the image
AB= h0= heigh of the object.
FP= f= focal length of the mirror.

From figure, In ΔBAP and ΔB'A'P are similar triangles. So,
(A'B')/AB = (A'P)/AP
(A'B')/AB = -v/u <---------- (i)


In ΔB'A'F and ΔMNF are similar triangles. So,
A'B'/MN = A'F/FN
Here, MN=AB and FN≃FP, we have
(A'B')/AB = (A'P+FP)/FP
(A'B')/AB = (-v+f)/f = -(v-f)/f <-------- (ii)
Equating equation (i) and equation (ii)
-v/u = -(v-f)/f
-vf = -uv+vf
Dividing both sides by uvf, we get
uv/uvf = (uf/uvf) + (vf/uvf)
1/f = 1/v + 1/u
therefore, 1/f = 1/u + 1/v
This is the expression of mirror formula.

Linear Magnification
It is defined as ratio of image distance to the object distance. It is denoted by 'm'.
i.e. m= v/u
Also, linear magnification is defined as the ratio of size of image to the size of object.
i.e. M= I/O (no unit.)

1. When m= 0 this means the image is point in size.
2. When m= 1 this means image distance is equal to the size of object.
3. When m < 1.
4. When m > 1.
5. m= +ve, the image is real.
6. m= -ve, the image is virtual.




In case of any problem ask me in qustions section!!!!

© 2020 onlinelearnal.com