Chapter 36, Problem 36.46P

A diverging lens has a focal length of magnitude 20.0 cm. (a) Locate 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) For each case, find the magnification.

To determine

The location of the image for given object distance.

Answer to Problem 36.46P

The location of the image for an object distance of (i) $40.0\text{cm}$, (ii) $20.0\text{cm}$, and (iii) $10.0\text{cm}$ are $13.3\text{cm}$, $10.0\text{cm}$ and $6.67\text{cm}$ in front of the lens respectively.

Explanation of Solution

Given: The magnitude of the focal length of the lens is $20.0\text{cm}$.

Write the equation for lens.

$\begin{array}{c}\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\\ q=\frac{pf}{p-f}\end{array}$ (1)

Here,

$p$ is the object distance.

$q$ is the image distance.

$f$ is the focal length of the lens.

The focal length is negative for diverging lens. The value of focal length is $-20.0\text{cm}$.

For case (i), the object distance is $40.0\text{cm}$.

Substitute $40.0\text{cm}$ for $p$ and $-20.0\text{cm}$ for $f$ in equation (1) to find $q$.

$\begin{array}{c}q=\frac{\left(40.0\text{cm}\right)\left(-20.0\text{cm}\right)}{\left(40.0\text{cm}\right)-\left(-20.0\text{cm}\right)}\\ =-13.3\text{cm}\end{array}$

Thus, the location of the image for an object distance of $40.0\text{cm}$ is $13.3\text{cm}$ in front of the lens.

For case (ii), the object distance is $20.0\text{cm}$.

Substitute $20.0\text{cm}$ for $p$ and $-20.0\text{cm}$ for $f$ in equation (1) to find $q$.

$\begin{array}{c}q=\frac{\left(20.0\text{cm}\right)\left(-20.0\text{cm}\right)}{\left(20.0\text{cm}\right)-\left(-20.0\text{cm}\right)}\\ =-10.0\text{cm}\end{array}$

Thus, the location of the image for an object distance of $20.0\text{cm}$ is $10.0\text{cm}$ in front of the lens.

For case (iii), the object distance is $10.0\text{cm}$.

Substitute $10.0\text{cm}$ for $p$ and $-20.0\text{cm}$ for $f$ in equation (1) to find $q$.

$\begin{array}{c}q=\frac{\left(10.0\text{cm}\right)\left(-20.0\text{cm}\right)}{\left(10.0\text{cm}\right)-\left(-20.0\text{cm}\right)}\\ =-6.67\text{cm}\end{array}$

The location of the image for an object distance of $10.0\text{cm}$ is $6.67\text{cm}$ in front of the lens.

Conclusion:

Therefore, the location of the image for an object distance of (i) $40.0\text{cm}$, (ii) $20.0\text{cm}$, and (iii) $10.0\text{cm}$ are $13.3\text{cm}$, $10.0\text{cm}$ and $6.67\text{cm}$ in front of the lens respectively.

To determine

Whether the image for the given object distance is real or virtual.

Answer to Problem 36.46P

The image for an object distance of (i) $40.0\text{cm}$, (ii) $20.0\text{cm}$, and (iii) $10.0\text{cm}$ is virtual for all the cases.

Explanation of Solution

Given: The magnitude of the focal length of the lens is $20.0\text{cm}$.

The image is real when the image distance is positive $\left(q>0\right)$ and the image is virtual when the image distance is negative $\left(q<0\right)$.

From the calculation of part (a) in case (i), the image distance is,

$q=-13.3\text{cm}$

Thus, the image distance is $-13.3\text{cm}$ which is negative due to which the image is virtual.

From the calculation of part (a) in case (ii), the image distance is,

$q=-10.0\text{cm}$

Thus, the image distance is $-10.0\text{m}$ which is negative due to which the image is virtual.

From the calculation of part (a) in case (iii), the image distance is,

$q=-6.67\text{cm}$

The image distance is $-6.67\text{cm}$ which is negative due to which the image is virtual.

Conclusion:

Therefore, the image for an object distance of (i) $40.0\text{cm}$, (ii) $20.0\text{cm}$, and (iii) $10.0\text{cm}$ is virtual for all the cases.

To determine

Whether the image for the given object distance is upright or inverted.

Answer to Problem 36.46P

The image for an object distance of (i) $40.0\text{cm}$, (ii) $20.0\text{cm}$, and (iii) $10.0\text{cm}$ is upright for all the cases.

Explanation of Solution

Given: The magnitude of the focal length of the lens is $20.0\text{cm}$.

Write the equation for magnification of the image.

$M=-\frac{q}{p}$ (2)

Here,

$M$ is the magnification of the image.

The image is upright when the magnification is positive $\left(M>0\right)$ and the image is inverted when the magnification is negative $\left(M<0\right)$.

For case (i), the object distance is $40.0\text{cm}$ and the image distance is $13.3\text{cm}$.

Substitute $-13.3\text{cm}$ for $q$ and $40.0\text{cm}$ for $p$ in equation (2) to find $M$.

$\begin{array}{c}M=-\frac{\left(-13.3\text{cm}\right)}{\left(40.0\text{cm}\right)}\\ =0.333\end{array}$

The magnification is $0.333$ which is positive due to which the image is upright.

For case (ii), the object distance is $20.0\text{cm}$ and the image distance is $-10.0\text{cm}$.

Substitute $-10.0\text{cm}$ for $q$ and $20.0\text{cm}$ for $p$ in equation (2) to find $M$.

$\begin{array}{c}M=-\frac{\left(-10.0\text{cm}\right)}{\left(20.0\text{cm}\right)}\\ =0.500\end{array}$

The magnification is $0.500$ which is positive due to which the image is upright.

For case (iii), the object distance is $10.0\text{cm}$ and the image distance is $-6.67\text{cm}$.

Substitute $-6.67\text{cm}$ for $q$ and $10.0\text{cm}$ for $p$ in equation (2) to find $M$.

$\begin{array}{c}M=-\frac{\left(-6.67\text{cm}\right)}{\left(10.0\text{cm}\right)}\\ =0.667\end{array}$

The magnification is $0.667$ which is positive due to which the image is upright.

Conclusion:

Therefore, the image for an object distance of (i) $40.0\text{cm}$, (ii) $20.0\text{cm}$, and (iii) $10.0\text{cm}$ is upright for all the cases.

To determine

The magnification for the given object distance.

Answer to Problem 36.46P

The magnification for an object distance of (i) $40.0\text{cm}$, (ii) $20.0\text{cm}$, and (iii) $10.0\text{cm}$ are $0.333$, $0.500$ and $0.667$ respectively.

Explanation of Solution

Given: The magnitude of the focal length of the lens is $20.0\text{cm}$.

From the calculation of part (c) in case (i), the magnification for an object distance of $40.0\text{cm}$ is,

$M=0.333$

Thus, the magnification is $0.333$.

From the calculation of part (c) in case (ii), the magnification for an object distance of $20.0\text{cm}$ is,

$M=0.500$

Thus, the magnification is $0.500$.

From the calculation of part (c) in case (iii), the magnification for an object distance of $10.0\text{cm}$ is,

$M=0.667$

The magnification is $0.667$.

Conclusion:

Therefore, the magnification for an object distance of (i) $40.0\text{cm}$, (ii) $20.0\text{cm}$, and (iii) $10.0\text{cm}$ are $0.333$, $0.500$ and $0.667$ respectively.