The refractive index μ of a transparent medium varies with wavelength λ of light as μ = A + ( B λ 2 ) , Where A and B are constants. Find the dimensional formula and SI units of A and B .

The refractive index μ of a transparent medium varies with the wavelength λ of light as μ = A + ( B λ 2 ) , Where A and B are constants. μ is dimensionless, therefore using the principle of Dimensional Homogeneity, A will dimensionless and has no unit. Similarly, ( B λ 2 ) is also dimensionless. On rearranging the terms, the dimension of B becomes to that of λ 2 . So the dimension of B becomes [ M 0 L 2 T 0 ] and its SI unit is m 2 .

The refractive index μ of a transparent medium varies with wavelength λ of light as μ = A + ( B λ 2 ) , Where A and B are constants. Find the dimensional formula and SI units of A and B .

The refractive index μ of a transparent medium varies with the wavelength λ of light as μ = A + ( B λ 2 ) , Where A and B are constants. μ is dimensionless, therefore using the principle of Dimensional Homogeneity, A will dimensionless and has no unit. Similarly, ( B λ 2 ) is also dimensionless. On rearranging the terms, the dimension of B becomes to that of λ 2 . So the dimension of B becomes [ M 0 L 2 T 0 ] and its SI unit is m 2 .

Find the dimension of the constant a , b , c and d in the relation v = a + b t + c d + t , where v is velocity and t is time.

The velocity is given by v = a + b t + c d + t Where a , b and c are constants. Using the principle of Dimensional Homogeneity, a will have dimension of velocity i.e. [ M 0 L 1 T - 1 ] ( c d + t ) will have dimensions of velocity. On rearranging the terms, the dimensions of c becomes to that of v t 1 . So the dimension of b becomes [ M 0 L 1 T 0 ] . ( b t ) will have dimensions of velocity. On rearranging the terms, the dimensions of b becomes to that of v t - 1 . So the dimensions of b becomes [ M 0 L 1 T - 2 ] The dimensions of d will have the dimension of time. The dimension of d will be [ M 0 L 0 T 1 ] .

Find the dimensions of the constant a × b in the relation E = ( b - x 2 ) a t , where E is the energy, x is the distance and t is the time.

The Energy of the particle E = ( b - x 2 ) a t Where a and b are constants and x is the distance . Using the principle of Dimensional Homogeneity, b will have dimension of x 2 i.e. [ M 0 L 2 T 0 ] ( a t ) will have dimension of E - 1 x 2 . On rearranging the terms, the dimension of a becomes to that of t - 1 E - 1 x 2 . So the dimension of a becomes [ T ] - 1 [ M 1 L 2 T - 2 ] - 1 [ L 2 ] = [ M - 1 L 0 T 1 ] The dimension of a × b = [ M 0 L 2 T 0 ] [ M - 1 L 0 T 1 ] = [ M - 1 L 2 T 1 ]

E M B I B E

ISC Physics Class XI Part 1>Chapter 5 - Dimensional Analysis>NUMERICALS>Q 3

MEDIUM

11th ICSE

IMPORTANT

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The refractive index $μ$ of a transparent medium varies with wavelength $λ$ of light as $μ = A + (\frac{B}{λ^{2}}),$ Where $A$ and $B$ are constants. Find the dimensional formula and SI units of $A$ and $B$ .

Important Questions on Dimensional Analysis

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11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 5 - Dimensional Analysis>NUMERICALS>Q 4

Find the dimension of the constant $a, b, c$ and $d$ in the relation $v = a + b t + \frac{c}{d + t}$ , where $v$ is velocity and $t$ is time.

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11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 5 - Dimensional Analysis>NUMERICALS>Q 5

Find the dimension of $\frac{a}{b}$ in the relation $v = a + b t,$ where $v$ is velocity, $t$ is time and $a$ and $b$ are constants.

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 5 - Dimensional Analysis>NUMERICALS>Q 6

Find the dimensions of the constant $a \times b$ in the relation $E = \frac{(b - x^{2})}{a t}$ , where $E$ is the energy, $x$ is the distance and $t$ is the time.