Chapter 23, Problem 49P

To determine

(a)

The distance in front of the sensor and the lens if the object to be photographed is $10.0m$ away.

Answer to Problem 49P

Solution:

$v=106.11mm$

Explanation of Solution

Given:

$105mm$ focal length lens is used to focus an image on the sensor of a camera. The maximum distance allowed between the lens and the sensor place is $132mm$. I.e.,

Object distance, $u=-10000mm$.

Image distance, $v=132mm$.

Focal length, $f=105mm$.

Formula used:

Lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$u$ and $v$ are lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

Calculation:

The distance in front of the sensor and the lens if the object to be photographed is $10m$ away is calculated using lens formula.

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

Here value of $u$ and $f$ are given, i.e., $u=-10000mm$ and $f=105mm$. Thus substituting them in formula we get-

$\begin{array}{l}\frac{1}{v}+\frac{1}{10000}=\frac{1}{105}\\ v=\frac{10000\times 105}{10000-105}\\ v=106.11mm\end{array}$

To determine

(b)

The distance in front of the sensor and the lens if the object to be photographed is $3.0m$ away.

Answer to Problem 49P

Solution:

$v=108.81mm$.

Explanation of Solution

Given:

$105mm$ focal length lens is used to focus an image on the sensor of a camera. The maximum distance allowed between the lens and the sensor place is $132mm$. I.e.,

Object distance, $u=-3000mm$.

Image distance, $v=132mm$.

Focal length, $f=105mm$.

Formula used:

Lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$u$ and $v$ are lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

Calculation:

The distance in front of the sensor and the lens if the object to be photographed is $3.0m$ away is calculated using lens formula.

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

Here value of $u$ and $f$ are given, i.e., $u=-3000mm$ and $f=105mm$. Thus substituting them in formula we get-

$\begin{array}{l}\frac{1}{v}+\frac{1}{3000}=\frac{1}{105}\\ v=\frac{3000\times 105}{3000-105}\\ v=108.81mm\end{array}$

To determine

(c)

The distance in front of the sensor and the lens if the object to be photographed is $1.0m$ away.

Answer to Problem 49P

Solution:

$v=117.32mm$.

Explanation of Solution

Given:

$105mm$ focal length lens is used to focus an image on the sensor of a camera. The maximum distance allowed between the lens and the sensor place is $132mm$. I.e.,

Object distance, $u=-1000mm$.

Image distance, $v=132mm$.

Focal length, $f=105mm$.

Formula used:

Lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$u$ and $v$ are lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

Calculation:

The distance in front of the sensor and the lens if the object to be photographed is 1.0m away is calculated using lens formula.

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

Here value of $u$ and $f$ are given, i.e., $u=-1000mm$ and $f=105mm$. Thus substituting them in formula we get-

$\begin{array}{l}\frac{1}{v}+\frac{1}{1000}=\frac{1}{105}\\ v=\frac{1000\times 105}{1000-105}\\ v=117.32mm\end{array}$

To determine

(d)

The closest object photographed sharply by the lens.

Answer to Problem 49P

Solution:

$u=-513.33mm$.

Explanation of Solution

Given:

$105mm$ focal length lens is used to focus an image on the sensor of a camera. The maximum distance allowed between the lens and the sensor place is $132mm$. I.e.,

Object distance, $u=-10000mm$.

Object distance, $u=-3000mm$.

Object distance, $u=-1000mm$.

Image distance, $v=132mm$.

Focal length, $f=105mm$.

Formula used:

Lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$u$ and $v$ are lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

Calculation:

The closest object photographed sharply by the lens is calculated using lens formula.

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

Here value of $v$ and $f$ are given, i.e., $v=132mm$ and $f=105mm$. Thus substituting them in formula we get-

$\begin{array}{l}\frac{1}{132}-\frac{1}{u}=\frac{1}{105}\\ \frac{1}{u}=\frac{1}{132}-\frac{1}{105}\\ u=\frac{132\times 105}{105-132}\\ u=-513.33mm\end{array}$