write the relation between R,N,F
Answers
Answer:MPN= Concave spherical mirror
P=Pole
F= Principle focus
C=Centre of curvature
PC=Principle axis
f= Focal length of mirror
R= Radius of curvature of mirror
AD= incident ray
EF=reflected ray
DC= Normal at point D
Angle of incidence=∠ADC
Angle of reflection=∠CDF
∠ADC=∠ CDF=θ(let)(By second law of reflection)
∠ADC=∠DCF=θ(alternate angles)
In ΔDCF,
∠CDF=∠DCF
ΔDCF is a isocsleles triangle
∴DF=PC
⎣
⎢
⎢
⎡
sidesopp. toequal.
angles in a triangle are
equal
⎦
⎥
⎥
⎤
If the aperture of mirror is very small i.e., point D is very near to point P then.
DF=PF(approx)
Hence, PF=FC
PE+PF=PF+FC
2PF=PC ...(i)
Applying sign convention,
PF=−f,PF=R
Put in eq. (1)
2(−f)=−R
R=2f
OR
f=
2
R
Hence, focal length of concave mirror is half of its radius of curvature.
For convex mirror:
Let, MPN=convex spherical mirror
P=Pole
F= Principle focus
C= Centure of curvature
PC= Principle axis
f= Focal length of mirror
R= Radius of curvature of mirror
AD= Incident ray
DB=Reflected ray
CE= Normal at point D
Angle of incidence=∠ADE
Angle of reflection=∠EDB
∠ADE=∠EDB=θ(let) (By 2
nd
law of reflection)
∠EDB=∠ FDC=θ(vertically opposite angles)
∠ADE=∠FCD=θ(corresponding angles)
ΔDCF is a isosceles triangle
∴DF=FC
⎣
⎢
⎢
⎡
Sides opp. to equal
angles in a triangle are
equal
⎦
⎥
⎥
⎤
If the aperture of mirror is very small i.e., point D is very near to point P then.
DF=PF
Hence, PF=FC
PF+PF=PF+FC .......(1)
Applying sign convetion, $$
PF=+f,PC=+R
Put in eq.(1)
R=2f
OR
f=
2
R
Hence focal length of concave mirror is half of its radius of curvature
solution
Explanation: