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 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 = 130cm, 100 cm, 90 cm and 80 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
Explanation
Slope = \(\frac{Δd^2/d^2}{ΔD^{-1}\) → \(frac{y_2 - y_1}{x_2 - x_1}\)
= \(\frac{0.4697 - 0.6638}{0.431 - 0.0254}\)
S = \(\frac{-0.1941}{0.0177}\) → -10,97
K = \(\frac{s}{4}\) → \(\frac{-10.97}{4}\)
= 2.74
Precautions:
(i) Lens was placed vertically on the lens holder. (ii) Avoided zero error in reading my metre rule.
(b)(ii) Differences between real and virtual images:
(i) Real image is formed on the screen while virtual image cannot be found on the screen. (ii) Real images are inverted while virtual images are erect.