Chapter 23, Problem 87GP

To determine

The length of image of the sun on a camera sensor with different focal lengths.

Answer to Problem 87GP

Solution:

Solution can be provided by magnification formula of lenses.

Explanation of Solution

Given:

Focal length of camera sensor is 35 mm

Diameter of Sun = 1∙ 4×10⁶ km = 1∙ 4×10⁹ m

Distance of sum from earth (object)= 1∙ 5×10⁸ km = 1∙ 5×10¹¹ m

Formula used:

Magnification of object by lenses is given by

$m=h_i/h_o=-d_i/d_o$, where $h_i$ and $h_o{}_{}$ is height of object and height of image and $d_i\text{and}{d}_o$ are distance of image and object.

Since Sun is very far from us then image would be formed at focal length, at f.

$d_i=f$ so, from the magnification formula

$h_i=\left(-d_i/d_o \right)h_o=\left(-f/d_o\right)h_o=-f{h}_o/d_o$

Calculation:

$f=35\text{ mm }=3\cdot 5\times {10}^{-2}\text{m}$

So, height of image of Sun,$h_i=-f{h}_o/d_o=\left(3\cdot 5\times {10}^{-2}\times 1\cdot 4\times {10}^9\right)/1\cdot 5\times {10}^{11}\text{m}$

$h_i=\left(4\cdot 9\times {10}^7\right)/1\cdot 5\times {10}^{11}\text{m }=3\cdot 27\times {10}^{-4}\text{m }=0\cdot 327\text{ mm}$

To determine

The length of image of the sun on a camera sensor with different focal lengths.

Answer to Problem 87GP

Solution:

Solution can be provided by magnification formula of lenses.

Explanation of Solution

Given:

Focal length of camera sensor is 50 mm

Diameter of Sun = 1∙ 4×10⁶ km = 1∙ 4×10⁹ m

Distance of sum from earth (object)= 1∙ 5×10⁸ km = 1∙ 5×10¹¹ m

Formula used:

Magnification of object by lenses is given by

$m=h_i/h_o=-d_i/d_o$, where $h_i\text{and}{h}_o{}_{}$ is height of object and height of image and $d_i\text{and}{d}_o$ are distance of image and object.

Since Sun is very far from us then image would be formed at focal length, at f.

$d_i=f$

So, from the magnification formula

$h_i=\left(-d_i/d_o \right)h_o=\left(-f/d_o\right)h_o=-f{h}_o/d_o$

Calculation:

$\begin{array}{l}f=50mm=5\cdot 0\times {10}^{-2}\text{m}\hfill \\ \text{So height of image of Sun},h_i=-f{h}_o/d_o=\left(5\cdot 0\times {10}^{-2}\times 1\cdot 4\times {10}^9\right)/1\cdot 5\times {10}^{11}\text{m}\hfill \\ h_i=\left(7\cdot 0\times {10}^7\right)/1\cdot 5\times {10}^{11}\text{m = 4}{\text{.67× 10}}^{\text{-4}}\text{m = 0}\text{.467 mm}\hfill \end{array}$

To determine

The length of image of the sun on a camera sensor with different focal lengths.

Answer to Problem 87GP

Solution:

Solution can be provided by magnification formula of lenses.

Explanation of Solution

Given:

Focal length of camera sensor is 105 mm

Diameter of Sun = 1∙ 4×10⁶ km = 1∙ 4×10⁹ m

Distance of sum from earth (object)= 1∙ 5×10⁸ km = 1∙ 5×10¹¹ m

Formula used:

Magnification of object by lenses is given by

$m=h_i/h_o=-d_i/d_o$, where $h_i\text{and}{h}_o{}_{}$ is height of object and height of image and $d_i\text{and}{d}_o$ are distance of image and object.

Since Sun is very far from us then image would be formed at focal length, at f.

$d_i=f$

So, from the magnification formula

$h_i=\left(-d_i/d_o \right)h_o=\left(-f/d_o\right)h_o=-f{h}_o/d_o$

Calculation:

$\begin{array}{l}{\text{f= 105 mm = 1×05×10}}^{\text{-1}}\text{m}\hfill \\ \text{So height of image of Sun},h_i=-f{h}_o/d_o=\left(1\cdot 05\times {10}^{-1}\times 1\cdot 4\times {10}^9\right)/1\cdot 5\times {\text{10}}^{\text{11}}\text{m}\hfill \\ {\text{h}}_{\text{i}}\text{= }\left({\text{1×47×10}}^{\text{8}}\right){\text{/ 1×5×10}}^{\text{11}}\text{m = 0}{\text{.98× 10}}^{\text{-3}}\text{m = 0}\text{.98 mm}\hfill \end{array}$