Question

A concave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) Find the location of the image for object distances of (i) 40.0 cm, (ii) 20.0 cm, and (iii) 10.0 cm. For each case, state whether the image is (b) real or virtual and (c) upright or inverted. (d) Find the magnification in each case.

Answer 1

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 :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

v is the image distance

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-40}$

v = -13.33 cm

Magnification of mirror is calculated as :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-13.33)}{(-40)}$    

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 :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

v is the image distance

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-20}$

v = -20 cm

Magnification of mirror is calculated as :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-20)}{(-40)}$    

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 :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

v is the image distance

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-10}$

v = infinity

Magnification of mirror is calculated as :

magnification = infinity

Hence, this is the required solution.

Answer 2

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.