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IB DP Physics Mock Exam SL Paper 1B Set 1 - 2025 Syllabus

IB DP Physics Mock Exam SL Paper 1B Set 1

Prepare for the IB DP Physics Mock Exam SL Paper 1B Set 1 with our comprehensive mock exam set 1. Test your knowledge and understanding of key concepts with challenging questions covering all essential topics. Identify areas for improvement and boost your confidence for the real exam

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Question 

Students are investigating refraction using a semi-circular glass block. Light from a ray box enters the curved surface and passes through the block, refracting at point P as it exits.
(a) Outline how the students can ensure that the light is not deflected at the curved surface.
The students vary the ray box position to collect data for determining the refractive index. They measure angles of incidence \(\theta_i\) and refraction \(\theta_r\) at P using a protractor.
 
 
 
 
 
 
 
 
 
 
(b) (i) For one measurement, the protractor is aligned as shown. State the value of \(\theta_r\) for this reading.

(ii) Complete the table by calculating \(\sin \theta_r\) for the row where \(\theta_r = 49^\circ\).

\(\theta_i\) (°)\(\theta_r\) (°)\(\sin \theta_i\)\(\sin \theta_r\)
10150.1740.259
20310.3420.515
25390.4230.629
30490.500 
35600.5740.866
40750.6430.966
They plot \(\sin \theta_r\) against \(\sin \theta_i\), add uncertainty bars for \(\sin \theta_i\) for the first and last points, and draw a best-fit line.
 
 
 
 
 
 
 
 
 
 
(c) (i) Determine the gradient of the best-fit line.
(ii) On the students’ graph, draw the line of maximum possible gradient.
(iii) Determine the refractive index of the glass and its absolute uncertainty.
▶️ Answer/Explanation

(a)
Ensure the incident ray is directed toward the centre P of the flat side so that it strikes along a radial line (angle of incidence = 0° at the curved surface).
\(\boxed{\text{Aim incident ray toward centre P of flat side}}\)

(b)(i)
From the protractor description, \(\theta_i\) is measured from the normal to the incident ray. The value is approximately \(48^\circ\)–\(49^\circ\). Since the table row for \(\theta_i = 30^\circ\) has \(\theta_r = 49^\circ\), the corresponding \(\theta_i\) for that measurement is \(30^\circ\).
\(\boxed{30^\circ}\)

(b)(ii)
For \(\theta_r = 49^\circ\), \(\sin 49^\circ \approx 0.7547 \approx 0.755\).
\(\boxed{0.755}\)

(c)(i)
Using two points on the best-fit line, e.g., (0.2, 0.3) and (0.8, 1.2):
\[ \text{Gradient} = \frac{1.2 – 0.3}{0.8 – 0.2} = \frac{0.9}{0.6} = 1.5 \]
\(\boxed{1.5}\)

(c)(ii)
Draw a straight line from the lower end of the first error bar (smallest \(\sin \theta_i\)) to the upper end of the last error bar (largest \(\sin \theta_r\)). This line should be steeper than the best-fit line.
\(\boxed{\text{Line drawn through extremes of error bars}}\)

(c)(iii)
The refractive index \(n = \text{gradient} = 1.5\).
Uncertainty from max/min gradients: If max gradient ≈ 1.7 and min ≈ 1.5 (or similar), \(\Delta n \approx 0.2\).
\(\boxed{n = 1.5 \pm 0.2}\)

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