Respuesta :
Answer:
Answered
Explanation:
The radius of curvature of the mirror R = 20 cm
then the focal length f = R/2 = 10 cm
(a) From mirror formula
1/f = 1/di + /1do
then the image distance
di = fd_o / d_o - f
= (10)(40) / 40-10
= 30.76 cm
since the image distance is positive so the image is real
ii) when the object distance d_0=20 cm
di = 10×20/ 20-10
= 20
Hence, the image must be real
iii)when the object distance d_0 = 10
di = 10×10 / 10-10 = ∞ (infinite)
the image will be formed at ∞
here also image will be real but diminished.
Explanation:
It is given that,
Radius of curvature of the mirror, R = 20 cm
So, focal length of the mirror, f = -10 cm
(i) Object distance, u = -40 cm
Using the mirror's formula as :
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
v is the image distance
[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-40}[/tex]
v = -13.33 cm
Magnification of mirror is calculated as :
[tex]m=\dfrac{-v}{u}[/tex]
[tex]m=\dfrac{-(-13.33)}{(-40)}[/tex]
m = -0.33
Since, the magnification is negative, image is real and inverted.
(ii) Object distance, u = -20 cm
Using the mirror's formula as :
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
v is the image distance
[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-20}[/tex]
v = -20 cm
Magnification of mirror is calculated as :
[tex]m=\dfrac{-v}{u}[/tex]
[tex]m=\dfrac{-(-20)}{(-40)}[/tex]
m = -0.5
Since, the magnification is negative, image is real and inverted.
(iii) Object distance, u = -10 cm
Using the mirror's formula as :
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
v is the image distance
[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-10}[/tex]
v = infinity
Magnification of mirror is calculated as :
magnification = infinity
Hence, this is the required solution.