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Science Answer the questions.[Phy-5M]

Light – Reflection and Refraction · CBSE STD 10 (CBSE - English Medium). 3 questions.

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Answer the questions.[Phy-5M]

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Question 15 Marks
Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.
Answer
Laws of refraction of light are:

  • The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
  • The ratio of sine of angle of incidence to the sine of angle of refraction is constant, for the light of a given wavelength and for the given pair of media. This law is also known as Snell’s law of refraction. The constant is also known as refractive index.

The ray diagram is as shown below:

In the glass slab, the emergent rays are parallel to the incident ray because the extant of bending of the ray of light at the opposite parallel faced of rectangular glass slab are equal and opposite, so that emergent ray is parallel to incident ray.

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Question 25 Marks
Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.
Answer
Perform an experiment to demonstrate that light bends from its path, when it falls obliquely on the surface of a glass slab. Also show that angle of incidence is about equal to the emergent angle.

  1. Take a glass slab and place it on a white sheet of paper fixed on a drawing board.
  2. Mark the boundary ABCD of the glass slab.
  3. Fix two pins P1 and P2, vertically on the drawing board such that line joining the pins is inclined to the edge AB of the glass slab (Figure).

  1. Now, look through the glass slab from opposite side so that the images of pins P1 and P2 are seen exactly in line. Fix two pins P3 and P4, vertically on the drawing board such that pins P3, P4 and image of P1 and P2 are exactly in one line.
  2. Remove the glass slab and all pins. Join the points P1 and P2 with a line and extend this line to touch the edge AB at P. Similarly join the points P3 and P4 with a line and extend the line to touch the edge DC at Q.
  3. Join points P and Q with a straight line.
  4. Draw normal NN’ at P on the edge AB and a normal N1N1' at Q on the edge DC.
  5. Measure $\angle\text{P}_2\text{PN}$ and $\angle\text{P}_3\text{QN}_1'$ using a protractor or dee.
This experiment shows that when light falls obliquely on a glass slab, it bends along PQ from its original path along PL.

Also $\angle\text{P}_2\text{PN}\approx\angle\text{P}_3\text{QN}_1'.$ That is, angle of incidence is about equal to the emergent angle.

Note: Distance between pins P1 and P2, and between P3 and P4 must be large.

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Question 35 Marks
A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under,
Position of candle = 12.0cm
Position of convex lens = 50.0cm
Position of the screen = 88.0cm
  1. What is the focal length of the convex lens?
  2. Where will the image be formed if he shifts the candle towards the lens at a position of 31.0cm?
  3. What will be the nature of the image formed if he further shifts the candle towards the lens?
  4. Draw a ray diagram to show the formation of the image in case (iii) as said above.
Answer
Let f be the focal length of the convex lens.

The distance of object should be measured from pole of the lens.

Distance of candle (or object) = Position of convex lens - Position of candle = 50 - 12 = 38cm.

Now, by sign convention, distance of candle (or object) = u = -38cm

Similarly, distance of candle’s image = position of the screen - position of convex lens = 88 - 50 = 38cm

By sign convention, distance of candle’s image = v = +38cm

  1. Using lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{f}}=\frac{1}{38}-\frac{1}{-38}=\frac{1}{19}$

$\Rightarrow\ \text{f}=19\text{cm}$

The focal length of the convex lens is 19cm.

  1. When the candle is shifted towards the lens at a position of 31.0cm.

Then, new object distance = position of convex lens c position of candle = 50 – 31 = 19

By sign convention, u = -19cm.

Now, focal length of the convex lens = 19cm.  It means, the candle lays at the focus of lens, hence its image is formed at infinity.

  1. When he further shifts the candle towards the lens. This means candle lies between optical centre and focus of convex lens, so, magnified, virtual and erect image of the candle will be formed.
  2. The ray diagram of image formation is given below:

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