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

Worked Solution & Example Answer:1. (a) Which ray diagram shows total internal reflection at an air and glass boundary? (1) (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

(b)(i) Using information from Figure 1, determine the magnification of the virtual image.

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Answer

To find the magnification of the virtual image, we use the formula:

magnification=height of imageheight of objectmagnification = \frac{height\ of\ image}{height\ of\ object}

Given the heights from Figure 1 (assuming the heights are labeled or can be measured from the diagram), substitute those values into the equation to find the magnification.

Step 2

(b)(ii) Describe one way the magnification of the image could be increased.

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Answer

One way to increase the magnification of the image is to use a different lens or replace the current lens with one that has a higher power or shorter focal length. This change enhances the curvature of the lens and increases the refractive index, leading to a larger virtual image.

Step 3

(c) Calculate the focal length, f, of the lens.

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Answer

To calculate the focal length, we can apply the lens formula:

1f=1a+1b\frac{1}{f} = \frac{1}{a} + \frac{1}{b}

Substituting the values:

  • a = 20 cm
  • b = 40 cm

We compute:

1f=120+140=2+140=340\frac{1}{f} = \frac{1}{20} + \frac{1}{40} = \frac{2 + 1}{40} = \frac{3}{40}

Thus,

f=40313.33 cmf = \frac{40}{3} \approx 13.33\ cm

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