Physics-
General
Easy

Question

A light ray travelling in a glass medium is incident on glass - air interface at an angle of incidence q . The reflected (R) and transmitted (T) intensities, both as function ofq , are plotted. The correct sketch is

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The correct answer is:

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A plane mirror is placed at the bottom of a tank containing a liquid of refractive index m . P is a small object at a height h above the mirror. An observer O, vertically above P, outside the liquid, observes P and its image in the mirror. The apparent distance between these two will be

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A ray of light travels in the way as shown in the figure. After passing through water, the ray grazes along the water air interface. The value of mg interms of ‘i’ is open parentheses mu subscript w end subscript equals 4 divided by 3 close parentheses

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A reflecting surface is represented by the equation x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent. A ray travelling in negative x-direction is directed towards positive y-direction after reflection from the surface at point P. Then co-ordinates of point P are

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If 3 Tan to the power of 4 space alpha minus 10 Tan squared space alpha plus 3 equals 0 then principal values of ' alpha ' are

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If Sin space open parentheses fraction numerator pi Cot space theta over denominator 4 end fraction close parentheses equals Cos space open parentheses fraction numerator pi Tan space theta over denominator 4 end fraction close parentheses and theta is in the first quadrant then theta =....

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If Tan squared space theta equals square root of 3 plus left parenthesis square root of 3 minus 1 right parenthesis Tan space theta and theta lies in open parentheses negative pi over 2 comma pi over 2 close parentheses then theta =

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If , thenTan space theta plus Sec space theta equals square root of 3 he principal value of open parentheses theta plus pi over 6 close parentheses is

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In a straight triangle A B C comma sin to the power of 4 invisible function application A plus sin to the power of 4 invisible function application B plus sin to the power of 4 invisible function application C equals

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The value of sum from k equals 1 to 100 of   sin space left parenthesis k x right parenthesis cos space left parenthesis 101 minus k right parenthesis x is equal to

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If alpha comma beta comma gamma text  are acute angles and  end text cos text end text theta equals sin space beta divided by sin space alpha comma cos space ϕ equals sin space gamma divided by sin invisible function application alpha and cos space left parenthesis theta minus ϕ right parenthesis equals sin space beta sin space gamma then the value of tan squared space alpha minus tan squared space beta minus tan squared space gamma is equal to

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Ifx sin space a plus y sin space 2 a plus z sin space 3 a equals sin space 4 a x sin space b plus y sin space 2 b plus z sin space 3 b equals sin space 4 b comma x sin space c plus y sin space 2 c plus z sin space 3 c equals sin space 4 c Then the roots of the equation t cubed minus open parentheses z over 2 close parentheses t squared minus open parentheses fraction numerator y plus z over denominator 4 end fraction close parentheses t plus open parentheses fraction numerator z minus x over denominator 8 end fraction close parentheses equals 0 comma a comma b comma c not equal to n pi are

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If equals fraction numerator 2 pi over denominator 7 end fraction,then tan alpha tan invisible function application 2 alpha plus tan invisible function application 2 alpha tan invisible function application 4 alpha plus tan invisible function application 4 alpha tan invisible function application alpha equals

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If the mapping f left parenthesis x right parenthesis equals a x plus b comma a less than 0 maps [-1,1] onto [0,2] then for all values of thetaA equals cos squared space theta plus sin to the power of 4 space theta is such that

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