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In your answer book copy the diagram on the right, which shows a light ray incident on the interface between glass and air - Leaving Cert Physics - Question c - 2016

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In your answer book copy the diagram on the right, which shows a light ray incident on the interface between glass and air. In your diagram, sketch (i) the refract... show full transcript

Worked Solution & Example Answer:In your answer book copy the diagram on the right, which shows a light ray incident on the interface between glass and air - Leaving Cert Physics - Question c - 2016

Step 1

Sketch (i) the refracted ray

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Answer

Begin by drawing the interface between the glass and air at a slight angle. Indicate the angle of incidence (the angle between the incident ray and the normal line) to be any angle less than the critical angle of 42° to ensure refraction occurs. From the point of incidence, draw the refracted ray bending away from the normal into the air. The angle of refraction can be determined using Snell's Law:

n1sin(θ1)=n2sin(θ2)n_1 \sin(\theta_1) = n_2 \sin(\theta_2)

where n1n_1 is the refractive index of glass (approximately 1.5), and n2n_2 is the refractive index of air (approximately 1.0).

To find the angle of refraction, if the angle of incidence is assumed to be 30°, for example:

1.5sin(30°)=1.0sin(θ2)1.5 \sin(30°) = 1.0 \sin(\theta_2)

From this, you can solve for θ2\theta_2.

Step 2

Sketch (ii) the weak reflected ray

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Answer

From the same point of incidence on the boundary, draw the weak reflected ray. This ray will travel back into the glass, at the same angle to the normal as the angle of incidence, thus following the law of reflection:

θi=θr\theta_{i} = \theta_{r}

where θi\theta_{i} is the angle of incidence and θr\theta_{r} is the angle of reflection. Ensure to label both the reflected and refracted rays clearly in the diagram.

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