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## BOARD QUESTIONS FROM WAVE OPTICS (CBSE/AISSCE)

Download the collection of questions asked in previous CBSE board exams from the chapter Wave Optics.

Practising the previously asked questions is a good idea to prepare for the exams.

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## Polarisation – Malus Law

Two polariods P1 and P2 are placed with their pass axes perpendicular to each other. Unpolarised light of intensity Io is incident on P1. A third polariod P3 is kept between P1 and P2 such that its pass axis makes an angle of 30 with that of P1. Determine the intensity of light transmitted through P1,P2 and P3.

Asked Rashmi

Answer

Appply Malus’ Law

The intensity of light from the first polaroid P1 is Io/2

From P3, the intensity is (Io/2) cos230 =(Io/2)(3/4)=3Io/8

From P2, the intensity is (3Io/8)cos260= (3Io/8)(1/4)= 3Io/32

## How to find the focal length of convex mirror using convex lens?

How to find the focal length of convex mirror using convex lens?

Asked Manisha

Answer:

We cannot use an optical method to determine the focal length of a convex mirror without using a convex (converging) lens since a convex mirror does not form a real image of an object in front of it or even parallel rays coming from infinity.

Therefore we have to use a convex mirror to determine the focal length of the convex mirror by an indirect method.

### STEP 1 DETERMINE THE ROUGH FOCAL LENGTH OF THE CONVEX LENS

First of all we take a convex lens and determine  its rough focal length by forming a real image of a distant object on a screen. The distance between the convex lens and the screen gives the rough focal length, since when the object is at infinity, the image formed by the convex lens is at its focus. This method is called the distant object method to estimate the focal length of a convex lens. The focal length of a concave mirror may also be estimated using this method.

Now, as we know the approximate focal length of the convex lens, keep an object- an optical pin or a lit candle in front of the convex lens at a distance around 2 times the rough focal length determined and place a screen on the other side so as to form a real image of the object on it. This adjustment is done so as to make the distances involved quite manageable.

When you get a clear and sharp image on the screen, mark the positions of the convex lens and the screen on the table.

Place the convex mirror in between the screen and the lens without disturbing the position of the candle and the lens.

Place the screen close to the candle. Carefully adjust the position of the convex mirror alone so that you get a sharp image on the screen which is now kept  at the position of and along with the candle. Mark the position of the convex mirror now.

The distance between the position of the convex mirror and the old position of the screen gives the radius of curvature of the convex mirror. (Since r = 2f; we can determine the focal length by dividing the distance with 2)

### Why the distance is r (2f)?

We know that when a ray of light is incident normally on the surface of a mirror, it retraces its path.

We also know that a line drawn from the centre of curvature of the mirror to the surface of a mirror is normal to the surface.

Therefore, when we are getting the image at the same position as the candle, it is formed by retracing of the rays, which means that the old position of the screen is at the position of the centre of curvature of the convex lens.

## Optics Concept map

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## Optics question bank

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## Optics value based questions

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## YDSE (Young’s Double Slit Experiment)

In the question we have a simple YDSE setup , there’s just the addition of a glass slab of thickness t and ,refractive index u, in front of the top slit ( S1). Now we submerge the entire setup into water (given refractive index u1). We now have to find the locations of the points, which are the points of constructive interference( max intensity).

I considered 3 approaches, first-

I calculated the optical path difference in water = t(u/u1 -1)- yd/D. where y is the reference coordinate of the maxima( any general position) from the original central maxima, d is the slit separation , and D is the distance from screen. For constructive interference path difference = n*wavelength( in water). Now wavelength in water= “wavelength in air”/u1 Therefore we get our answer.

Approach two:

I found out the path difference in water and multiplied it by u1 , to get the path difference in air , and I got the same result when I equated this path difference to- n*wavelength(in air).

Approach three- I found the path difference in air – t(u-1)-yd/D and divided this by u1, to get path difference in water, and equated this to n* wavelength (water). This time around , I did not get the answer.

Can anyone explain the difference between the second and third approaches. And also please do tell me, if I have committed some error.

Posted by Garvit Sharma

## Focal length of convex mirror using convex lens

It is given that the image formed by the convex mirror should be at radius of curvature . Why it should?

Asked

Answer:

I guess that the question is based on the experiment to determine the focal length of a convex mirror using convex lens.

Initially a convex lens is used to obtain a real image of a candle. (or pin if you are doing the experiment using the optical bench) Then a convex mirror is kept in such a way that the image is obtained on the same position as the object. This happens when the rays are incident normal to the mirror. The rays of light incident normal to the mirror are directed towards the centre of curvature. Since the image was formed by the rays which are now reflected normally, the distance between the convex mirror and the original position of image gives the radius of curvature.

See the diagram for illustration.

## if a biconvex lens is made with different materials vertically and horizontally then how many images will be formed?

if a biconvex lens is made with different materials vertically and horizontally then how many images will be formed?

Asked Mahathi

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