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Question-45747




Question Number 45747 by rahul 19 last updated on 16/Oct/18
Answered by ajfour last updated on 16/Oct/18
p_(1cm) ^� = m(((p_1 ^� +p_2 ^� )/(2m))−(p_1 ^� /m))=((p_2 ^� −p_1 ^� )/2)  λ_(1cm) =(h/(∣p_(1cm) ^� ∣)) = ((2h)/( (√(p_1 ^2 +p_2 ^2 ))))        = (2/( (√((1/((h/p_1 )^2 ))+(1/((h/p_2 )^2 ))))))      = (2/( (√((1/λ_1 ^2 )+(1/λ_2 ^2 ))))) = ((2λ_1 λ_2 )/( (√(λ_1 ^2 +λ_2 ^2 )))) .
$$\bar {{p}}_{\mathrm{1}{cm}} =\:{m}\left(\frac{\bar {{p}}_{\mathrm{1}} +\bar {{p}}_{\mathrm{2}} }{\mathrm{2}{m}}−\frac{\bar {{p}}_{\mathrm{1}} }{{m}}\right)=\frac{\bar {{p}}_{\mathrm{2}} −\bar {{p}}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\lambda_{\mathrm{1}{cm}} =\frac{{h}}{\mid\bar {{p}}_{\mathrm{1}{cm}} \mid}\:=\:\frac{\mathrm{2}{h}}{\:\sqrt{{p}_{\mathrm{1}} ^{\mathrm{2}} +{p}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{2}}{\:\sqrt{\frac{\mathrm{1}}{\left({h}/{p}_{\mathrm{1}} \right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({h}/{p}_{\mathrm{2}} \right)^{\mathrm{2}} }}} \\ $$$$\:\:\:\:=\:\frac{\mathrm{2}}{\:\sqrt{\frac{\mathrm{1}}{\lambda_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{\lambda_{\mathrm{2}} ^{\mathrm{2}} }}}\:=\:\frac{\mathrm{2}\lambda_{\mathrm{1}} \lambda_{\mathrm{2}} }{\:\sqrt{\lambda_{\mathrm{1}} ^{\mathrm{2}} +\lambda_{\mathrm{2}} ^{\mathrm{2}} }}\:. \\ $$
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_(1cm) =((p_2 −p_1 )/2)  ,  then how ∣p_(1cm) ∣=((√(p_1 ^2 +p_2 ^2 ))/2) ?
$$\:{p}_{\mathrm{1}{cm}} =\frac{{p}_{\mathrm{2}} −{p}_{\mathrm{1}} }{\mathrm{2}}\:\:, \\ $$$${then}\:{how}\:\mid{p}_{\mathrm{1}{cm}} \mid=\frac{\sqrt{{p}_{\mathrm{1}} ^{\mathrm{2}} +{p}_{\mathrm{2}} ^{\mathrm{2}} }}{\mathrm{2}}\:? \\ $$
Commented by ajfour last updated on 16/Oct/18
since p_1 ^�  and p_2 ^�  are at right angles,  ∣p_2 ^� −p_1 ∣=(√(p_1 ^2 +p_2 ^2 ))   just like     say  p_1 ^� = 3i  , p_2 ^� =4j^�   , then  ∣p_1 ^� −p_2 ^� ∣=∣3i^� −4j^� ∣=(√(3^2 +4^2 )) = 5 .
$${since}\:\bar {{p}}_{\mathrm{1}} \:{and}\:\bar {{p}}_{\mathrm{2}} \:{are}\:{at}\:{right}\:{angles}, \\ $$$$\mid\bar {{p}}_{\mathrm{2}} −{p}_{\mathrm{1}} \mid=\sqrt{{p}_{\mathrm{1}} ^{\mathrm{2}} +{p}_{\mathrm{2}} ^{\mathrm{2}} }\:\:\:{just}\:{like} \\ $$$$\:\:\:{say}\:\:\bar {{p}}_{\mathrm{1}} =\:\mathrm{3}{i}\:\:,\:\bar {{p}}_{\mathrm{2}} =\mathrm{4}\hat {{j}}\:\:,\:{then} \\ $$$$\mid\bar {{p}}_{\mathrm{1}} −\bar {{p}}_{\mathrm{2}} \mid=\mid\mathrm{3}\hat {{i}}−\mathrm{4}\hat {{j}}\mid=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }\:=\:\mathrm{5}\:. \\ $$

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