Chapter 24, Problem 90QAP

To determine

(a)

The final image due to the lens-mirror combination.

Answer to Problem 90QAP

The image formed by the mirror is located $750cm$ front of the mirror.

Explanation of Solution

Given:

A small, Plano-concave lens $f_A=-45cm$

Radius of curvature $+60m$

Focal length of spherical mirror is equal to the radius of curvature divided in half

  $f_B=30cm$$$

Formula used:

  $\frac{1}{f_A}=\frac{1}{d_{O,A}}+\frac{1}{d_{1,A}}$$$

Calculation:

  $\frac{1}{f_A}=\frac{1}{d_{O,A}}+\frac{1}{d_{1,A}}$

  $\begin{array}{l}\frac{1}{d_{1,A}}=\frac{1}{f_A}-\frac{1}{d_{O,A}}\\ \frac{1}{d_{1,A}}=\frac{d_{O,A}-f_A}{f_A{d}_{O,A}}\\ d_{1,A}=\frac{f_A{d}_{O,A}}{d_{O,A}-f_A}\end{array}$

  $\begin{array}{l}{d}_{1,A}=\frac{\left(-45\right)\left(15\right)}{\left(15\right)-\left(-45\right)}\\ d_{1,A}=\frac{-675}{60}\\ d_{1,A}=-11.25cm\end{array}$

Image formed by the mirror:

The image formed by lens is located $11.25cm$ behind the lens,

  $d_{O,B}=11.25+20=31.25cm$

  $\begin{array}{l}\frac{1}{d_{O,B}}+\frac{1}{d_{1,B}}=\frac{1}{f_B}\\ \frac{1}{d_{1,B}}=\frac{1}{f_B}-\frac{1}{d_{O,B}}\\ \frac{1}{d_{1,B}}=\frac{d_{O,B}-f_B}{f_B{d}_{O,B}}\\ d_{1,B}=\frac{f_B{d}_{O,B}}{d_{O,B}-f_B}\\ d_{1,B}=\frac{\left(30\right)\left(31.25\right)}{31.25-30}\\ d_{1,B}=750cm\end{array}$

Conclusion:

The image formed by the mirror is located $750cm$ front of the mirror.

To determine

(b)

Find out final image is the virtual or real.

Answer to Problem 90QAP

The image is inverted $m_{total}<0$.

Explanation of Solution

Given:

A small, Plano-concave lens $f_A=-45cm$

Radius of curvature $+60m$

Focal length of spherical mirror is equal to the radius of curvature divided in half

  $f_B=30cm$$$

Formula used:

  $m_{total}=m_A{m}_B$$$

Calculation:

Since $d_{1,B}>0$, the image is real.

Total magnification:

  $\begin{array}{l}{m}_{total}=m_A{m}_B\\ m_{total}=\left(-\frac{d_{1,A}}{d_{O,A}}\right)\left(-\frac{d_{1,B}}{d_{O,b}}\right)\\ m_{total}=\left(\frac{-11.25}{15}\right)\left(\frac{750}{31.25}\right)\\ m_{total}=-18\end{array}$

Conclusion:

The image is inverted $m_{total}<0$. $$

To determine

(c)

Calculate the final image due to the new lens-mirror combination from repeating part (a) and (b) when concave mirror is replaced by a convex mirror of same radius.

Answer to Problem 90QAP

  $d_{1,B}<0$, the image created by the mirror is virtual.

The image is upright $m_{total}>0$.

Explanation of Solution

Given:

A small, Plano-concave lens $f_A=-45cm$

Radius of curvature $+60m$

Convex mirror of same magnitude radius of curvature have focal length is:

  $f_B=-30cm$$$

Formula used:

  $m_{total}=m_A{m}_B$$$

Calculation:

  $\begin{array}{l}\frac{1}{d_{O,B}}+\frac{1}{d_{1,B}}=\frac{1}{f_B}\\ \frac{1}{d_{1,B}}=\frac{1}{f_B}-\frac{1}{d_{O,B}}\\ \frac{1}{d_{1,B}}=\frac{d_{O,B}-f_B}{f_B{d}_{O,B}}\\ d_{1,B}=\frac{f_B{d}_{O,B}}{d_{O,B}-f_B}\\ d_{1,B}=\frac{\left(-30\right)\left(31.25\right)}{31.25-\left(-30\right)}\\ d_{1,B}=-15.3cm\end{array}$

Since $d_{1,B}<0$, the image created by the mirror is virtual.

Total magnification:

  $\begin{array}{l}{m}_{total}=m_A{m}_B\\ m_{total}=\left(-\frac{d_{1,A}}{d_{O,A}}\right)\left(-\frac{d_{1,B}}{d_{O,b}}\right)\\ m_{total}=\left(\frac{-11.25}{15}\right)\left(\frac{-15.3}{31.25}\right)\\ m_{total}=0.367\end{array}$

Conclusion:

  $d_{1,B}<0$, the image created by the mirror is virtual.

The image is upright $m_{total}>0$. $$