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1. (a) Which ray diagram shows total reflection at an air and glass boundary? (b) Figure 1 is a ray diagram for a converging lens used as a magnifying glass - Edexcel - GCSE Physics - Question 1 - 2022 - Paper 1

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1.-(a)-Which-ray-diagram-shows-total-reflection-at-an-air-and-glass-boundary?----(b)-Figure-1-is-a-ray-diagram-for-a-converging-lens-used-as-a-magnifying-glass-Edexcel-GCSE Physics-Question 1-2022-Paper 1.png

1. (a) Which ray diagram shows total reflection at an air and glass boundary? (b) Figure 1 is a ray diagram for a converging lens used as a magnifying glass. (... show full transcript

Worked Solution & Example Answer:1. (a) Which ray diagram shows total reflection at an air and glass boundary? (b) Figure 1 is a ray diagram for a converging lens used as a magnifying glass - Edexcel - GCSE Physics - Question 1 - 2022 - Paper 1

Step 1

Describe one way the magnification of the image could be increased.

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Answer

To increase the magnification of the virtual image, you can use a different lens or replace the existing lens with one of higher power, which typically has a shorter focal length. Alternatively, moving the object further away from the lens will also increase the magnification.

Step 2

Calculate the focal length, f, of the lens.

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Answer

Given that the distance from the object to the lens is a=20 extcma = 20 \, ext{cm} and from the image to the lens is b=40 extcmb = 40 \, ext{cm}, we can apply the formula:

1f=a+ba×b\frac{1}{f} = \frac{a + b}{a \times b}

Substituting the values, we calculate:

1f=20+4020×40=60800=340\frac{1}{f} = \frac{20 + 40}{20 \times 40} = \frac{60}{800} = \frac{3}{40}

Therefore, the focal length is:

f=403≈13.33 cmf = \frac{40}{3} \approx 13.33 \, \text{cm}

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