Chapter 25, Problem 39P

To determine

The magnification of the binoculars when focussed on a object placed at a finite distance from the objective.

Answer to Problem 39P

Solution:

The magnification of the binoculars when focussed on an object placed at a distance of 4.0 m away from the objective is found to be $7\times$

Explanation of Solution

Binoculars function in a similar manner as a telescope, but they return an upright image as they use prisms to invert the image formed by the lenses. The expression for angular magnification remains the same as it is for a telescope, since the prisms do not produce magnification; they merely turn the refracted rays from the objective.

Angular magnification is given by the expression,

$M=\frac{\theta \text{'}}{\theta }$

Here, $\theta \text{'}$ is the angle subtended by the object at the eye when using the device (eyepiece lens)and $\theta$ is the angle it subtends in the absence of the lens.

If the object is at infinity, the image of the object is formed at the focus of the objective but for objects not at infinity, the image is formed at a distance $v_0$ from the object.

Therefore, if h is the height of the image formed by the objective (Fig 25-20 of the text book), then for the object at a finite distance from the objective, the magnification is given by,

$\begin{array}{c}M=\frac{\theta \text{'}}{\theta }=\frac{h/f_e}{h/v_0}\\ =\frac{v_0}{f_e}\end{array}$

Here, the focal length of the eyepiece is $f_e$.

When the eye is relaxed, then the magnification m of the binoculars is given by the expression,

$m=\frac{f_o}{f_e}$

The focal length of the eyepiece can be calculated using the above expression.

The image distance $v_0$ can be calculated by using the mirror equation.

$\frac{1}{f_o}=\frac{1}{u_o}+\frac{1}{v_o}$

Given:

The magnification of the binoculars $m=6.5$

Focal length of the objective $f_o=26\text{ cm}=0.26\text{ m}$

Distance of the object from the objective $u_o=4.0\text{ m}$

Formula used:

$m=\frac{f_o}{f_e}$

$\frac{1}{f_o}=\frac{1}{u_o}+\frac{1}{v_o}$

$M=\frac{v_0}{f_e}$

Calculation:

Calculate the focal length of the eyepiece lens $f_e$ using the given values of magnification and the focal length of the objective.

$\begin{array}{c}{f}_e=\frac{f_o}{m}\\ =\frac{0.26\text{ m}}{6.5}\\ =0.04\text{ m}\end{array}$

Use the given values of object distance and focal length of the objective to calculate the distance $v_0$ at which the image is formed from the objective, using the mirror equation $\frac{1}{f_o}=\frac{1}{u_o}+\frac{1}{v_o}$.

According to sign conventions the focal length of a convex lens is positive, for real objects, the object distance is positive and when a real image is formed, the image distance is also positive.

$\begin{array}{c}\frac{1}{v_o}=\frac{1}{f_o}-\frac{1}{u_o}\\ =\frac{1}{0.26\text{ m}}-\frac{1}{4\text{ m}}=\frac{4.26}{1.04}\end{array}$

The image is formed at a distance $v_o=0.278\text{ m}$ from the objective.

Use the calculated values of $f_e$ and $v_0$ to calculate the magnification of the binoculars.

$\begin{array}{c}M=\frac{v_0}{f_e}\\ =\frac{0.278\text{ m}}{0.04\text{ m}}\\ =6.95\\ =7\times \end{array}$