resolving power

The resolving power may be increased by increasing the reflection coefficient. This can be done by silvering, or by appropriate dielectric layers - just as blooming a lens can reduce reflection, so appropriate coating can increase reflectivity.

The basic considerations which apply to the definition of the resolving power are the same as those that Rayleigh used in his definition of the criterion for the resolution of two slits, namely that the peaks have to be far enough apart compared with their widths that there is a discernible dip in intensity between them This definition is not suitable for the present case, as we do not have a zero in the diffraction pattern. Instead, we use a definition due to Taylor, which states that the peaks are resolvable if their separate intensity curves intersect at their half-intensity positions. Near $\theta$ = 0 ,

$$\delta$$ = $${4 \pi \over \lambda}$$d,

and differentiating and using the derivative as an approximation to the ratio of differences

$${{\rm d} \delta \over {\rm d} \lambda}$$ = - $${4 \pi d \over \lambda^2}$$ $$\approx$$ $${\Delta \delta \over \Delta \lambda}$$.

If the peaks are to be separated, the difference in $\delta$ between the peaks corresponding to the different wavelengths must be greater than double the half-width at half-height of each peak,

$${4 \over \sqrt{F}}$$ < $${4 \pi d \over \lambda^2}$$$$\Delta$$$$\lambda$$.

This gives a resolving power

$${\lambda \over (\Delta\lambda)_{\rm min}}$$ = $${\pi d \over \lambda}$$$$\sqrt{F}$$.

A reasonable criterion for the resolving power is

$${\lambda \over \Delta\lambda_{\rm min}}$$ = $${\pi d \over \lambda}$$$$\sqrt{F}$$.

For example, if R = 0.9 , d = 10 mm , $\lambda$ = 500 nm , we find that

F = $${4 \times 0.9 \over (1 - 0.9)^2}$$ = 360,

and then

$${\lambda \over \Delta\lambda_{\rm min}}$$ = 1.2 x 10⁶

which is significantly larger than is achieved with most diffraction gratings (recall our previous result for a grating, which had a resolving power of 2000).