Question

Please help me complete section 3 with data points provided

Answer 1

Given:

To find:

Magnification of an optical system with uncertainties.

Step-by-step explanation:

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

`M=(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(0.05)/(12.8)*100\% \\ \\ 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(0.05)/(9.7)*100\% \\ \\ x=9.7\pm0.52\%..........(3) \end{gathered}`

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

`\begin{gathered} M=(12.8\pm0.39\%)/(9.7\pm0.52\%) \\ \\ M=(12.8)/(9.7)\pm(0.39+0.52)\% \\ \\ M=1.32\pm0.91\% \\ \\ M=1.32\pm(0.91)/(100)*1.32 \\ \\ 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=(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(0.05)/(403)*100\% \\ \\ 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(0.05)/(210)*100 \\ \\ o=210\pm0.0238\%..........(6) \end{gathered}`

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

`\begin{gathered} M=(403\pm0.0124\%)/(210\pm0.0238\%) \\ \\ M=(403)/(210)\pm(0.0124+0.0238)\% \\ \\ M=1.92\pm0.0362\% \\ \\ M=1.92\pm(0.0362)/(100)*1.92 \\ \\ 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.