Q93P - University Physics with Modern Physics

Question

A convex spherical mirror with a focal length of magnitude 24.0 cm is placed 20.0 cm to the left of a plane mirror. An object 0.250 cm tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?

Step-by-Step Solution

Answer

The images are 7.06 cm and -13.3cm from the mirror.

The height of the images are 0.177cm and 0.111cm.

Step 1:Relationship between object and image distance and height.

Object’s distance is s and image distance is s’.

The relationship is $\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{\mathit{f}}$ 

Object’s height is y and image height is y’,

Therefore $y\text{'}=-\frac{s\text{'}}{s}y$ 

Step 2:Calculation of the distance of images from the mirrors.

Image 1.

$$\begin{gathered} s=\frac{20\mathit{c}\mathit{m}}{2} \\ =10\mathit{c}\mathit{m} \end{gathered}$$ 

Using the relationship of distance between object and mirror,

$$\begin{gathered} s\text{'}=\frac{\mathbf{s}\mathbf{f}}{\mathbf{s}\mathbf{-}\mathbf{f}} \\ =\frac{10*\left(-24\right)}{10-\left(-24\right)} \\ =-7.06\mathit{c}\mathit{m} \end{gathered}$$ 

Image 2.

The focal length will be the same whereas the distance of the object will become 30cm.

Using the relationship of distance between object and mirror,

$$\begin{gathered} s\text{'}=\frac{\mathbf{sf}}{\mathbf{s}\mathbf{-}\mathbf{f}} \\ =\frac{30*\left(-24\right)}{30-\left(-24\right)} \\ =-13.3\mathit{c}\mathit{m} \end{gathered}$$ 

Step 3: Calculation of the height of the image.

The height of the object is given as 0.25 cm.

Using the relationship between the height of image and object ,

Image 1.

$$\begin{gathered} y\text{'}=-\frac{s\text{'}}{s}y \\ =-\left(\frac{-7.06}{10}\right)*0.25 \\ =0.177\mathit{c}\mathit{m} \end{gathered}$$ 

 

Image 2.

 $$\begin{gathered} y\text{'}=-\frac{s\text{'}}{s}y \\ =-\left(\frac{-13.3}{30}\right)*0.25 \\ =0.111\mathit{c}\mathit{m} \end{gathered}$$