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(C) For a plane transmission grating,the condition for the $n$th order diffraction maxima is given by $d \sin 	heta = n \lambda$,where $d$ is the gra

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the images $C, B$ and $A$ will shift towards $O$

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(C) For a plane transmission grating,the condition for the $n$th order diffraction maxima is given by $d \sin 	heta = n \lambda$,where $d$ is the grating element,$	heta$ is the angle of diffraction,and $\lambda$ is the wavelength of light.From the relation $\sin 	heta = \frac{n \lambda}{d}$,it is clear that for a fixed order $n$ and grating constant $d$,the angle of diffraction $	heta$ is directly proportional to the wavelength $\lambda$ (i.e.,$	heta \propto \lambda$ for small angles).When the source is replaced by another source of shorter wavelength,the value of $\lambda$ decreases. Consequently,$\sin 	heta$ decreases,which means the angle of diffraction $	heta$ for all orders $(n=1, 2, 3)$ decreases.The direct image at $O$ (the $0$th order) remains stationary because $\sin 	heta = 0$ for $n=0$,regardless of the wavelength. However,the diffracted images $A, B$,and $C$ will move closer to the central maximum $O$ as their respective diffraction angles decrease.Therefore,the images $C, B$,and $A$ will shift towards $O$.

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