Question Number 45747 by rahul 19 last updated on 16/Oct/18
Answered by ajfour last updated on 16/Oct/18
$$\bar {{p}}_{\mathrm{1}{cm}} =\:{m}\left(\frac{\bar {{p}}_{\mathrm{1}} +\bar {{p}}_{\mathrm{2}} }{\mathrm{2}{m}}−\frac{\bar {{p}}_{\mathrm{1}} }{{m}}\right)=\frac{\bar {{p}}_{\mathrm{2}} −\bar {{p}}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\lambda_{\mathrm{1}{cm}} =\frac{{h}}{\mid\bar {{p}}_{\mathrm{1}{cm}} \mid}\:=\:\frac{\mathrm{2}{h}}{\sqrt{{p}_{\mathrm{1}} ^{\mathrm{2}} +{p}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{2}}{\sqrt{\frac{\mathrm{1}}{\left({h}/{p}_{\mathrm{1}} \right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({h}/{p}_{\mathrm{2}} \right)^{\mathrm{2}} }}} \\ $$$$\:\:\:\:=\:\frac{\mathrm{2}}{\sqrt{\frac{\mathrm{1}}{\lambda_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{\lambda_{\mathrm{2}} ^{\mathrm{2}} }}}\:=\:\frac{\mathrm{2}\lambda_{\mathrm{1}} \lambda_{\mathrm{2}} }{\sqrt{\lambda_{\mathrm{1}} ^{\mathrm{2}} +\lambda_{\mathrm{2}} ^{\mathrm{2}} }}\:. \\ $$
Commented by rahul 19 last updated on 16/Oct/18
Thank you sir ! ☺️�� plss also try Q. 45746 , 45748
Commented by rahul 19 last updated on 16/Oct/18
$$\:{p}_{\mathrm{1}{cm}} =\frac{{p}_{\mathrm{2}} −{p}_{\mathrm{1}} }{\mathrm{2}}\:\:, \\ $$$${then}\:{how}\:\mid{p}_{\mathrm{1}{cm}} \mid=\frac{\sqrt{{p}_{\mathrm{1}} ^{\mathrm{2}} +{p}_{\mathrm{2}} ^{\mathrm{2}} }}{\mathrm{2}}\:? \\ $$
Commented by ajfour last updated on 16/Oct/18
$${since}\:\bar {{p}}_{\mathrm{1}} \:{and}\:\bar {{p}}_{\mathrm{2}} \:{are}\:{at}\:{right}\:{angles}, \\ $$$$\mid\bar {{p}}_{\mathrm{2}} −{p}_{\mathrm{1}} \mid=\sqrt{{p}_{\mathrm{1}} ^{\mathrm{2}} +{p}_{\mathrm{2}} ^{\mathrm{2}} }\:\:\:{just}\:{like} \\ $$$$\:\:\:{say}\:\:\bar {{p}}_{\mathrm{1}} =\:\mathrm{3}{i}\:\:,\:\bar {{p}}_{\mathrm{2}} =\mathrm{4}\hat {{j}}\:\:,\:{then} \\ $$$$\mid\bar {{p}}_{\mathrm{1}} −\bar {{p}}_{\mathrm{2}} \mid=\mid\mathrm{3}\hat {{i}}−\mathrm{4}\hat {{j}}\mid=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }\:=\:\mathrm{5}\:. \\ $$