You have been provided with a ray box, a converging lens, a lens holder, a screen, a metre rule, and half- metre rule. Use the diagram above as a guide to p...
You have been provided with a ray box, a converging lens, a lens holder, a screen, a metre rule, and half- metre rule. Use the diagram above as a guide to perform the experiment.
(i) Determine the approximate focal length f, of the lens by focusing a distant object on the screen.
(ii) Place the ray box and the screen such that the distance between the illuminated cross-Wire and the screen, \(D= 150\ \text{cm}\).
(iii) Place the lens at a position L where a sharp mage of the cross-Wire Is obtained on the screen Note L.
(iv) Move the lens at a position L, to 0btain another sharp image of the cross-wire on the screen. Note L
(V) Measure the distance, d. between \(L_1\) and \(L_2\).
(Vi) Evaluate \(D^2\): \(d^2\) and \(D^2 - d^2\).
(vii) Repeat the procedure for four other values of \(D = 130\text{cm}, 100\text{ cm}, 90\text{ cm}\) and \(80\text{ cm}\). in each case.evaluate \(D^2\); \(d^2\) and \(D^2 - d^2\).
(viii) Tabulate the result
(ix) Plot a graph with \(D^2 - d^2\) on the vertical axis and \(D\) on the horizontal axis.
(x) Determine the r values of D axis and Determine the slopes, S, of the graph.
(xi) Evaluate \(k = \frac{s}{4}\)
(xii) State two precautions taken to ensure accurate results.
(bi) Distinguish between a virtual image and. plain image?
(ii) With the aid of a ray diagram, explain how a converging lens produces a Virtual image
Displacement (Bessel) method for the focal length of a converging lens
(i) Approximate focal length. With the ray box removed, a distant object (a window across the room) is focused sharply on the screen. The distance from the lens to the screen is measured with the metre rule:
\[ f \approx 15\ \text{cm} \]
Principle for parts (ii)-(viii). The illuminated cross-wire (object) and the screen are kept a fixed distance \(D\) apart. For any \(D>4f\) there are two positions of the lens, \(L_1\) and \(L_2\), that each throw a sharp image on the screen. Their separation is \(d=L_2-L_1\), and
The procedure of parts (iii)-(vii) is carried out for \(D=150,\,130,\,100,\,90\) and \(80\ \text{cm}\). The lens positions \(L_1,\,L_2\) are read off the metre rule (object taken as the zero of the scale), and the quantities \(D^{2}\), \(d^{2}\) and \(D^{2}-d^{2}\) are evaluated.
(viii) Table of results
\(D\) (cm)
\(L_1\) (cm)
\(L_2\) (cm)
\(d=L_2-L_1\) (cm)
\(D^{2}\) (cm\(^2\))
\(d^{2}\) (cm\(^2\))
\(D^{2}-d^{2}\) (cm\(^2\))
150
16.91
133.09
116.19
22500
13500
9000
130
17.31
112.69
95.39
16900
9100
7800
100
18.38
81.62
63.25
10000
4000
6000
90
19.02
70.98
51.96
8100
2700
5400
80
20.00
60.00
40.00
6400
1600
4800
(ix) Graph. \(D^{2}-d^{2}\) is plotted on the vertical axis against \(D\) on the horizontal axis. The points lie on a straight line passing through the origin.
Straight line through the origin; slope S = 60 cm, so k = S/4 = 15 cm = f.
(x) Intercept and slope. The line passes through the origin, so its intercept on the \(D\)-axis is
\[ D\text{-intercept}=0. \]
Taking two well-separated points on the line of best fit, \((80,\,4800)\) and \((150,\,9000)\):
(xi) Evaluate \(k\). Since \(D^{2}-d^{2}=4fD\), the slope \(S=4f\), hence
\[ k=\frac{S}{4}=\frac{60}{4}=15\ \text{cm}. \]
This equals the focal length of the lens and agrees with the value \(f\approx 15\ \text{cm}\) obtained in part (i).
(xii) Two precautions.
The lens was mounted vertically and its centre kept in line with the cross-wire and the screen, all at the same height on the bench.
Each image was focused as sharply as possible and the eye positioned squarely over the metre rule to avoid parallax and zero error when reading the lens positions.
(b)(i) Real image versus virtual image
Real image
Virtual image
Formed by the actual intersection of refracted rays.
Formed where the refracted rays only appear to come from when produced backwards.
Can be caught (focused) on a screen.
Cannot be caught on a screen.
Inverted relative to the object.
Erect (upright) relative to the object.
(b)(ii) How a converging lens produces a virtual image
The object is placed between the lens and its principal focus, that is at an object distance \(u
A ray travelling parallel to the principal axis is refracted so that it passes through the far principal focus \(F'\).
A ray directed through the optical centre \(C\) of the lens passes straight through undeviated.
After refraction these two rays diverge; they never meet on the far side. When they are produced backwards (broken lines) they meet on the same side as the object, forming an image that is virtual, erect and magnified.
Ray diagram: object inside the focal length of a converging lens gives a virtual, erect, magnified image (broken lines show the rays produced backwards).
Displacement (Bessel) method for the focal length of a converging lens
(i) Approximate focal length. With the ray box removed, a distant object (a window across the room) is focused sharply on the screen. The distance from the lens to the screen is measured with the metre rule:
\[ f \approx 15\ \text{cm} \]
Principle for parts (ii)-(viii). The illuminated cross-wire (object) and the screen are kept a fixed distance \(D\) apart. For any \(D>4f\) there are two positions of the lens, \(L_1\) and \(L_2\), that each throw a sharp image on the screen. Their separation is \(d=L_2-L_1\), and
The procedure of parts (iii)-(vii) is carried out for \(D=150,\,130,\,100,\,90\) and \(80\ \text{cm}\). The lens positions \(L_1,\,L_2\) are read off the metre rule (object taken as the zero of the scale), and the quantities \(D^{2}\), \(d^{2}\) and \(D^{2}-d^{2}\) are evaluated.
(viii) Table of results
\(D\) (cm)
\(L_1\) (cm)
\(L_2\) (cm)
\(d=L_2-L_1\) (cm)
\(D^{2}\) (cm\(^2\))
\(d^{2}\) (cm\(^2\))
\(D^{2}-d^{2}\) (cm\(^2\))
150
16.91
133.09
116.19
22500
13500
9000
130
17.31
112.69
95.39
16900
9100
7800
100
18.38
81.62
63.25
10000
4000
6000
90
19.02
70.98
51.96
8100
2700
5400
80
20.00
60.00
40.00
6400
1600
4800
(ix) Graph. \(D^{2}-d^{2}\) is plotted on the vertical axis against \(D\) on the horizontal axis. The points lie on a straight line passing through the origin.
Straight line through the origin; slope S = 60 cm, so k = S/4 = 15 cm = f.
(x) Intercept and slope. The line passes through the origin, so its intercept on the \(D\)-axis is
\[ D\text{-intercept}=0. \]
Taking two well-separated points on the line of best fit, \((80,\,4800)\) and \((150,\,9000)\):
(xi) Evaluate \(k\). Since \(D^{2}-d^{2}=4fD\), the slope \(S=4f\), hence
\[ k=\frac{S}{4}=\frac{60}{4}=15\ \text{cm}. \]
This equals the focal length of the lens and agrees with the value \(f\approx 15\ \text{cm}\) obtained in part (i).
(xii) Two precautions.
The lens was mounted vertically and its centre kept in line with the cross-wire and the screen, all at the same height on the bench.
Each image was focused as sharply as possible and the eye positioned squarely over the metre rule to avoid parallax and zero error when reading the lens positions.
(b)(i) Real image versus virtual image
Real image
Virtual image
Formed by the actual intersection of refracted rays.
Formed where the refracted rays only appear to come from when produced backwards.
Can be caught (focused) on a screen.
Cannot be caught on a screen.
Inverted relative to the object.
Erect (upright) relative to the object.
(b)(ii) How a converging lens produces a virtual image
The object is placed between the lens and its principal focus, that is at an object distance \(u
A ray travelling parallel to the principal axis is refracted so that it passes through the far principal focus \(F'\).
A ray directed through the optical centre \(C\) of the lens passes straight through undeviated.
After refraction these two rays diverge; they never meet on the far side. When they are produced backwards (broken lines) they meet on the same side as the object, forming an image that is virtual, erect and magnified.
Ray diagram: object inside the focal length of a converging lens gives a virtual, erect, magnified image (broken lines show the rays produced backwards).
