TEST OF PRACTICAL KNOWLEDGE QUESTION Determine and record the approximate focal length f\(_{o}\) of the concave mirror provided. Arrange the ray box, the mi...

(a) Determination of the focal length of the concave mirror

The approximate focal length obtained by focusing a distant object sharply on a screen is:

\[f_o=15.0\text{ cm}\]

For each object distance \(b\), the screen was adjusted until a sharp image was obtained. The image distance \(a\) was then measured and \(L=\dfrac{a}{b}\) calculated.

Graph of \(L\) against \(a\):

Using two well-separated points on the straight line, \((a_1,L_1)=(24.0\text{ cm},0.60)\) and \((a_2,L_2)=(60.0\text{ cm},3.00)\):

\[s=\frac{\Delta L}{\Delta a}=\frac{3.00-0.60}{60.0-24.0}=\frac{2.40}{36.0}=0.0667\text{ cm}^{-1}\]

\[s^{-1}=\frac{1}{0.0667}=15.0\text{ cm}\]

Therefore, \(\boxed{f_o=s^{-1}=15.0\text{ cm}}\).

Precautions:

1. The mirror, ray box and screen were kept collinear and upright.
2. The eye was placed directly above the metre rule scale to avoid parallax error when taking readings.

(b)(i) Characteristics of the image

For the concave mirror, \(f=15\text{ cm}\) and \(u=10\text{ cm}\). Using the Cartesian sign convention, \(f=-15\text{ cm}\) and \(u=-10\text{ cm}\):

\[\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\]

\[\frac{1}{-15}=\frac{1}{-10}+\frac{1}{v}\]

\[\frac{1}{v}=\frac{1}{30},\qquad v=+30\text{ cm}\]

The positive value of \(v\) shows that the image is formed \(30\text{ cm}\) behind the mirror and is virtual. Its magnification is:

\[m=-\frac{v}{u}=-\frac{30}{-10}=+3\]

Thus, the image is virtual, erect and magnified three times.

(b)(ii) How \(f_o\) was obtained

The concave mirror was directed towards a distant object and a screen was moved in front of it until a sharp image was formed. The distance between the pole of the mirror and the screen at sharp focus was measured. This distance is the approximate focal length \(f_o\).