 Okay, let us move to next, these two, all right, you are not able to solve the first one, 20 second, see when the ray will pass through a glass slab, the emergent ray will be parallel to the incident ray, so this ray has to be parallel to that, okay, similarly here also, this ray which will come out will be parallel, okay, this ray is parallel to that and this ray will be parallel to this, okay, and hence the angle remains alpha only, that is option 2 is correct, okay, what about 23rd, see in 23rd if you have equal widths then intensity coming out from both the slits will be I0 and I0, okay, if you increase the width of one of the slits to double, the intensity of one of the slits becomes 4 I0, okay, because area actually defines, area defines how much will be the intensity, so if you in, if you double it, intensity becomes 4 times, so 4 I0 and I0, so this is I1 and this is I2, alright, now you will see that the maximum possible intensity is root I1 plus root I2 the whole square which will come out to be 9 I0, okay, and minimum possible intensity is root I1 minus root I2 the whole square, okay, which is I0, so earlier the minimum intensity was 0 and maximum was 4 I0, now minimum is I0 and maximum is 9 I0, so both, maximum and minima, both of the intensities are increasing, alright, any doubt on these two questions, anything, alright, others, what are you getting, so if the person is able to see the object then the light from the object should reach the person, right, and light will follow law of reflection, so it will get reflected from let us say edges, from these two edges it will get reflected, then it will travel like this, so if this ray and that ray the person catches then he will be able to see the object, but if he goes beyond this point, if it suppose is here then he will not be able to see the, see this source or object, because this is the maximum distance up to which the light from the source can reach on this vertical line, fine, so you just need to find out what is the distance between this point and that point, so that is the answer, okay, so if I draw let us say this line and that line, okay, so due to symmetry you will see that this distance is D, okay, and then you will see that you know this is D, so this is D by 2, okay, and this distance will also be D by 2 because these two triangles are congruent, alright, and then you will see that 1, 2, 1, 2, 3 and 1, 4, 5, so triangle 1, 2, 3 is similar to triangle 1, 4, 5, right, and 1 to 5, distance 1 to 5 is 2 times the distance 1 to 2, so if you use ratio of sides, this length 4 to 5 length will be 2 times of 2 to 3, so if 2 to 3 is D by 2, this has to be D and similarly this distance will also come out to be D, fine, so the total distance will be D plus D plus D which will be 3D, that is why option number 4 is correct over here, fine, alright, so I think alright, so break time is over, fine, so let me upload new question, try solving this question, it refers to a figure, so I will just draw the figure over here, outside refractive index is N2 and inside refractive index is N1, okay, now try this, so if rate comes out from CD then it has to undergo total internal reflection on AD, okay, so that's the hint you can use, so N2 is more than N1, right, so it will bend away from the normal, so it has to at least graze the surface in order to reach CD, so angle of refraction over here should be 90 degrees, so this angle let's say this is theta has to be critical angle, should I solve this, okay, any other answer, alright, so let's see, this angle let us say is R, okay, so N2 sin of alpha should be equal to N1 sin of R, okay, so sin of R is equal to N2 by N1 sin of alpha, okay, or R is equal to sin inverse N2 minus N1 sin of alpha, fine, so if R is this, the angle of incidence at the surface AD has to be 90 minus R because this is right angle triangle, right, so theta has to be equal to pi by 2 minus R, now R is sin inverse of that, so this will become cos inverse of N2 by N1 sin of alpha, okay, so this is theta, okay, now we need to find the maximum value of alpha such that the ray comes out, okay, so if you increase the value of alpha beyond alpha max then cos inverse of that will come out to be less, okay, if this bracket term increases, cos inverse will decrease, so theta will decrease if you increase alpha, fine, so lesser the alpha better it is, okay, now at an angle when it just undergoes total internal reflection or there is this critical angle, in that case you will have N1 sin of theta which is N1 sin of cos inverse N2 by N1 sin of alpha this should be equal to N2, wait, N1 is more than N2, so this light will actually bend towards the normal, so R is less than alpha max, so this ray diagram is wrong but the way we have to do this whatever we have done is correct because we have not used the fact that R is less or more than alpha, okay, so this is N2 sin of 90 degrees, fine, so we will get sin of cos inverse N2 by N1 sin of alpha is equal to N2 by N1, alright, so then you rearrange this term you will get alpha to be equal to sin inverse of this, so that is why option number one is correct over here, okay, so lot of inverse technometric function usage is there in this particular question but if you just move ahead in a logical manner you will arrive at the correct answer, okay, so I am moving to next question, okay, what refracted ray will pass through CD without TIR, see we are assuming that AD is very large, the length AD is large, okay, so if it has some angle with the horizontal it has to cross the AD, so that is the assumption we are making, okay, is it clear, okay, now solve 26 and 27, two beams of light having intensity is I and 4I interfere to produce a fringe pattern, the phase difference is pi by 2 at A and pi at B, the difference between resultant intensity is at A and B we have to find out, so again this is direct application of the formula, resultant intensity is I1 plus I2 plus 2 under root I1 I2 cos of phase difference, okay, so initially when phase difference is pi by 2 intensity is I plus 4I that is 5I and when the phase difference is pi then it will be 5I plus 2 times under root of 4I2 cos of pi is minus 1, so this is I, so difference in IA and IB is 4I, alright, what about 27, for 27 there is a Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when wavelength of 6600 nanometer is used, now wavelength is changed you need to find number of fringes observed in the same segment, now first of all you should know that all the fringes will have the same width in Young's double slit experiment, right and the fringe width is given as lambda D by D and the segment length will be same in both the cases whether you use 600 or 400 the segment length is L only, so number of fringes will be L divided by beta, right, so let's say initially beta 1 is the fringe width and number of fringes is N1, so N1 is equal to L by beta 1, okay, so this will be L divided by beta 1 is lambda 1 D by D, so small D will go up, fine, this is N1 and N2 will be similarly L D divided by lambda 2 capital D, right, so this is equation 1 and that is equation 2, so if I divide these two I will get N1 by N2 to be equal to lambda 2 by lambda 1, okay, so N2 will be equal to N1 times lambda 1 by lambda 2, all right, so this will be 12 into lambda 1 which is 600 divided by 400, so we will have 6 by 4, 3 by 2, so it will be 18, so that is why option 2 is correct over here. Any doubts on these two questions? Anything?