Question Number 52593 by Tinkutara last updated on 10/Jan/19
Commented by Tinkutara last updated on 10/Jan/19
b part please
Answered by ajfour last updated on 11/Jan/19
$$\left(\mu_{\mathrm{2}} −\mu_{\mathrm{1}} \right){t}={d}\mathrm{sin}\:\theta_{\mathrm{0}} \\ $$$$\frac{{y}_{\mathrm{0}} }{{D}}\:=\:\mathrm{tan}\:\theta_{\mathrm{0}} \approx\:\mathrm{sin}\:\theta_{\mathrm{0}} \:=\:\frac{\left(\mu_{\mathrm{2}} −\mu_{\mathrm{1}} \right){t}}{{d}} \\ $$$${this}\:{gives}\:\:{y}_{\mathrm{0}} \:\approx\:\mathrm{1}.\mathrm{25}{cm} \\ $$$${y}_{\mathrm{1}} =\:{y}_{\mathrm{0}} +\frac{\lambda{D}}{{d}}\:=\:\mathrm{1}.\mathrm{25}{cm}\pm\mathrm{0}.\mathrm{49}{mm} \\ $$$$\:\:\:{i}\:{was}\:{wrong},\:{and}\:{i}\:{am}\:{still}\:{wrong}, \\ $$$${i}\:{shall}\:{try}\:{again}.. \\ $$$$ \\ $$
Commented by Tinkutara last updated on 11/Jan/19
But answers given are (a) 4.9×10^-4 m (b) 0.021 cm on one side and 0.028 cm on other side
Commented by Tinkutara last updated on 11/Jan/19
But Sir I also got the same answers as yours, then where is the mistake?