Q10E - University Physics with Modern Physics

Question

You hold a spherical salad bowl 60cm in front of your face with the bottom of the bowl facing you. The bowl is made of polished metal with a 35-cm radius of curvature. (a) Where is the image of your 5.0-cm tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

Step-by-Step Solution

Answer

a) The image of 5 cm tall nose will be located 13.55 cm inside the mirror.

b) The image’s size is 1.129 cm, it is erect and virtual.

Step 1: Object image relation for spherical mirror and Lateral Magnification

Object image relation for spherical mirror is given by:

                                                                      $\frac{\mathbf{1}}{\mathbf{u}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{v}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{R}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

Where, u= object distance from the vertex of the mirror,

u is (+) when the object is in front of the reflecting surface of mirror,

v is image distance from the vertex of the mirror,

v is (+) when the image is in front of the reflecting surface of mirror,

R is the radius of curvature of the mirror, f= focal length,

R is (+) in concave mirror and (-) in convex mirror,

u and v are negative if the case is opposite to what is mentioned above

Remember that the focal length of a mirror depends only on the curvature of the mirror and not on the material from which the mirror is made because the formation of the image results from rays reflected from the surface of the material. 

Apply: The above rule can be applied on either concave and convex mirrors or even plane mirror when we consider the radius to be infinity. The equation gives one the exact location of an image by knowing the radius or the focal length of a mirror and the location of the object. In addition, the sign for the distance of the image can be negative if the image in the opposite direction of the outgoing rays which means the image is not real image but is virtual one and is located inside the mirror.

Lateral magnification for spherical message is given by:

$\mathbf{m}\mathbf{=}\frac{\mathbf{‐}\mathbf{v}}{\mathbf{u}}\mathbf{=}\frac{{\mathbf{h}}_{\mathbf{i}}}{{\mathbf{h}}_{\mathbf{o}}}$

where, m is the magnification,

u is object distance from the vertex of the mirror,

u is (+) when the object is in front of the reflecting surface of mirror,

v is image distance from the vertex of the mirror,

v is (+) when the image is in front of the reflecting surface of mirror,

${\mathit{h}}_{\mathbf{i}}$ is the height of the image, ${\mathit{h}}_{\mathbf{0}}$ is the height of the object,

m is (+) when the image is erect and (-) when the image is inverted

Step 2: Determine location of image

a) $\frac{1}{u}+\frac{1}{v}=\frac{2}{R}=\frac{1}{f}$

The mirror is convex mirror, so

$$\begin{gathered} R=-35cm\left(given\right) \\ u=60cm \\ \frac{1}{v}=\frac{2}{R}-\frac{1}{u}=\frac{2u-R}{Ru} \\ v=\frac{Ru}{2u-R}=\frac{-35\times 60}{2\times 60-\left(-35\right)}=-13.55cm \end{gathered}$$

The image is $-13.55cm$ inside the mirror.

Step 3: Determine Lateral Magnification

b) $$\begin{gathered} m=\frac{‐v}{u}=\frac{h_i}{h_o}=\frac{-\left(-13.55cm\right)}{60cm}=0.2258, \\ {h}_i=0.2258\times h_o=0.225\times 5cm=1.129cm \end{gathered}$$

m is positive, so the image is erect.

v is negative, so the image is inside the mirror.

Therefore, the image is erect and virtual and size is 1.129 cm.