Question

Please help me complete section 3 with data points provided

Answer 1

Given:

To find:

Magnification of an optical system with uncertainties.

Explanation:

The formula for the magnification of an optical system is given as:

$M=\frac{x^{\prime}}{x}..........(1)$

The uncertain value of x' is:

$x^{\prime}=x^{\prime}\pm\sigma_{x^{\prime}}$

Substituting the values in the above equation, we get:

$x^{\prime}=12.8\pm0.05$

The uncertainty in the measurement of x' can be expressed in percentages as:

$\begin{gathered} x^{\prime}=12.8\pm\frac{0.05}{12.8}\times100\% \\ \\ x^{\prime}=12.8\pm0.39\%..........(2) \end{gathered}$

The uncertain value of x is:

$x=x\pm\sigma_x$

Substituting the values in the above equation, we get:

$x=9.7\pm0.05$

The uncertainty in the measurement of x can be expressed in percentages as:

$\begin{gathered} x=9.7\pm\frac{0.05}{9.7}\times100\% \\ \\ x=9.7\pm0.52\%..........(3) \end{gathered}$

Substituting values from equations (2) and (3) in equation (1), we get:

$\begin{gathered} M=\frac{12.8\pm0.39\%}{9.7\pm0.52\%} \\ \\ M=\frac{12.8}{9.7}\pm(0.39+0.52)\% \\ \\ M=1.32\pm0.91\% \\ \\ M=1.32\pm\frac{0.91}{100}\times1.32 \\ \\ M=1.32\pm0.012 \end{gathered}$

Thus, from the above equation, we get:

$M\pm\sigma_M=1.32\pm0.012$

The formula for the expected value of the magnification is given as:

$M=\frac{i}{o}..........(4)$

Here, i is the image position and o is the object position.

In the given data,

The image position is: t = 403

The object position is: a = 210

The image position with uncertain value is given as:

$i=t\pm\sigma_t$

Substituting the values in the above equation, we get:

$i=403\pm0.05$

The uncertainty in the measurement of image position can be expressed in percentages as:

$\begin{gathered} i=403\pm\frac{0.05}{403}\times100\% \\ \\ i=403\pm0.0124\%..........(5) \end{gathered}$

The object position with uncertain value is given as:

$o=a\pm\sigma_a$

Substituting the values in the above equation, we get:

$o=210\pm0.05$

The uncertainty in the measurement of the object position can be expressed in percentages as:

$\begin{gathered} o=210\pm\frac{0.05}{210}\times100 \\ \\ o=210\pm0.0238\%..........(6) \end{gathered}$

Substituting values from equations (5) and (6) in equation (4), we get:

$\begin{gathered} M=\frac{403\pm0.0124\%}{210\pm0.0238\%} \\ \\ M=\frac{403}{210}\pm(0.0124+0.0238)\% \\ \\ M=1.92\pm0.0362\% \\ \\ M=1.92\pm\frac{0.0362}{100}\times1.92 \\ \\ M=1.92\pm0.0007 \end{gathered}$

From the above equation, we get:

$M\pm\sigma_M=1.92\pm0.0007$

Final answer:

The magnification of the optical system is:

$M\pm\sigma_M=1.32\pm0.012$

The expected value of the magnification is:

$M\pm\sigma_M=1.92\pm0.0007$

When both the values are compared, we see that these values do not agree within their uncertainties.