(C) The number of waves $N$ in a medium of thickness $d$ and refractive index $\mu$ is given by $N = \frac{d}{\lambda_m}$,where $\lambda_m = \frac{\la

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(C) The number of waves $N$ in a medium of thickness $d$ and refractive index $\mu$ is given by $N = \frac{d}{\lambda_m}$,where $\lambda_m = \frac{\lambda_0}{\mu}$ is the wavelength in the medium.Thus,$N = \frac{d \cdot \mu}{\lambda_0}$.Since the number of waves is the same for both media,we have:$\frac{d_g \cdot \mu_g}{\lambda_0} = \frac{d_w \cdot \mu_w}{\lambda_0}$$d_g \cdot \mu_g = d_w \cdot \mu_w$Given $d_g = 4 \,cm$,$\mu_g = \frac{5}{3}$,and $\mu_w = \frac{4}{3}$.Substituting the values: $4 \times \frac{5}{3} = x \times \frac{4}{3}$$\frac{20}{3} = \frac{4x}{3}$$4x = 20$$x = 5 \,cm$.

Class 12 Physics Ray Optics and Optical Instruments Refraction through Plane Surface and Glass Slab

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$A$ glass slab of thickness $4 \,cm$ contains the same number of waves as in $x \,cm$ of water column when both are traversed by the same monochromatic light. If the refractive indices of glass and water for that light are $\frac{5}{3}$ and $\frac{4}{3}$ respectively,the value of $x$ will be

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A
$\frac{9}{20} \,cm$
B
$\frac{5}{4} \,cm$
C
$5 \,cm$
D
$\frac{20}{9} \,cm$

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Ray Optics and Optical Instruments

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Refraction through Plane Surface and Glass Slab

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$A$ glass slab of thickness $4 \,cm$ contains the same number of waves as in $x \,cm$ of water column when both are traversed by the same monochromatic light. If the refractive indices of glass and water for that light are $\frac{5}{3}$ and $\frac{4}{3}$ respectively,the value of $x$ will be

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$A$ microscope is focused on a point object and then its objective is raised through a height of $2 \ cm$. If a glass slab of refractive index $1.5$ is placed over this point object such that it is focused again,the thickness of the glass slab is......$cm$

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