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Question Number 45747 by rahul 19 last updated on 16/Oct/18

Answered by ajfour last updated on 16/Oct/18

$${\overline{p}}_{1cm}=m(\frac{{\overline{p}}_1+{\overline{p}}_2}{2m}-\frac{{\overline{p}}_1}{m})=\frac{{\overline{p}}_2-{\overline{p}}_1}{2}$$$${\lambda }_{1cm}=\frac{h}{\mid {\overline{p}}_{1cm}\mid }=\frac{2h}{\sqrt{p_1^2+p_2^2}}$$$$=\frac{2}{\sqrt{\frac{1}{{\left(h/p_1\right)}^2}+\frac{1}{{\left(h/p_2\right)}^2}}}$$$$=\frac{2}{\sqrt{\frac{1}{{\lambda }_1^2}+\frac{1}{{\lambda }_2^2}}}=\frac{2{\lambda }_1{\lambda }_2}{\sqrt{{\lambda }_1^2+{\lambda }_2^2}}.$$

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

$$p_{1cm}=\frac{p_2-p_1}{2},$$$$thenhow\mid p_{1cm}\mid =\frac{\sqrt{p_1^2+p_2^2}}{2}?$$

Commented by ajfour last updated on 16/Oct/18

$$since{\overline{p}}_{1}and{\overline{p}}_{2}areatrightangles,$$$$\mid {\overline{p}}_2-p_1\mid =\sqrt{p_1^2+p_2^2}justlike$$$$say{\overline{p}}_1=3i,{\overline{p}}_2=4\hat{j},then$$$$\mid {\overline{p}}_1-{\overline{p}}_2\mid =\mid 3\hat{i}-4\hat{j}\mid =\sqrt{3^2+4^2}=5.$$

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