Menu Close

Question-33452

Tinku Tara 0 Comments
Pin




Question Number 33452 by ajfour last updated on 16/Apr/18
Commented by ajfour last updated on 16/Apr/18
Find length of shadow s of a stick of length l when source of light is a point source at height H above ground.
$${Find}\:{length}\:{of}\:{shadow}\:{s}\:{of}\:{a}\: \\ $$$${stick}\:{of}\:{length}\:\boldsymbol{{l}}\:{when}\:{source} \\ $$$${of}\:{light}\:{is}\:{a}\:{point}\:{source}\:{at} \\ $$$${height}\:{H}\:{above}\:{ground}. \\ $$
Commented by ajfour last updated on 19/Apr/18
question for you, mrW Sir...
$${question}\:{for}\:{you},\:{mrW}\:{Sir}… \\ $$$$ \\ $$
Answered by MrW3 last updated on 12/Jun/18
Commented by MrW3 last updated on 12/Jun/18
x_1 =0, y_1 =H x_2 =u, y_2 =v x_3 =u+l cos α, y_3 =v+l sin α eqn. of line 12: ((x−0)/(y−H))=((u−0)/(v−H)) ⇒ x=(u/(v−H))(y−H) point 4: x_4 =(u/(v−H))(0−H)=((uH)/(H−v)) eqn. of line 13: ((x−0)/(y−H))=((u+l cos α−0)/(v+l sin α−H)) ⇒x=((u+l cos α)/(v+l sin α−H))(y−H) point 5: x_5 =((u+l cos α)/(v+l sin α−H))(0−H)=(((u+l cos α)H)/(H−v−l sin α)) s=x_5 −x_4 =(((u+l cos α)H)/(H−v−l sin α))−((uH)/(H−v)) =(((u+l cos α)(H−v)−u(H−v−l sin α))/((H−v−l sin α)(H−v)))×H ⇒s=(((H−v) cos α+u sin α)/((H−v−l sin α)(H−v)))×lH or ⇒s=(((H−y) cos α+x sin α)/((H−y−l sin α)(H−y)))×lH
$${x}_{\mathrm{1}} =\mathrm{0},\:{y}_{\mathrm{1}} ={H} \\ $$$${x}_{\mathrm{2}} ={u},\:{y}_{\mathrm{2}} ={v} \\ $$$${x}_{\mathrm{3}} ={u}+{l}\:\mathrm{cos}\:\alpha,\:{y}_{\mathrm{3}} ={v}+{l}\:\mathrm{sin}\:\alpha \\ $$$${eqn}.\:{of}\:{line}\:\mathrm{12}: \\ $$$$\frac{{x}−\mathrm{0}}{{y}−{H}}=\frac{{u}−\mathrm{0}}{{v}−{H}} \\ $$$$\Rightarrow\:{x}=\frac{{u}}{{v}−{H}}\left({y}−{H}\right) \\ $$$${point}\:\mathrm{4}: \\ $$$${x}_{\mathrm{4}} =\frac{{u}}{{v}−{H}}\left(\mathrm{0}−{H}\right)=\frac{{uH}}{{H}−{v}} \\ $$$${eqn}.\:{of}\:{line}\:\mathrm{13}: \\ $$$$\frac{{x}−\mathrm{0}}{{y}−{H}}=\frac{{u}+{l}\:\mathrm{cos}\:\alpha−\mathrm{0}}{{v}+{l}\:\mathrm{sin}\:\alpha−{H}} \\ $$$$\Rightarrow{x}=\frac{{u}+{l}\:\mathrm{cos}\:\alpha}{{v}+{l}\:\mathrm{sin}\:\alpha−{H}}\left({y}−{H}\right) \\ $$$${point}\:\mathrm{5}: \\ $$$${x}_{\mathrm{5}} =\frac{{u}+{l}\:\mathrm{cos}\:\alpha}{{v}+{l}\:\mathrm{sin}\:\alpha−{H}}\left(\mathrm{0}−{H}\right)=\frac{\left({u}+{l}\:\mathrm{cos}\:\alpha\right){H}}{{H}−{v}−{l}\:\mathrm{sin}\:\alpha} \\ $$$$ \\ $$$${s}={x}_{\mathrm{5}} −{x}_{\mathrm{4}} =\frac{\left({u}+{l}\:\mathrm{cos}\:\alpha\right){H}}{{H}−{v}−{l}\:\mathrm{sin}\:\alpha}−\frac{{uH}}{{H}−{v}} \\ $$$$=\frac{\left({u}+{l}\:\mathrm{cos}\:\alpha\right)\left({H}−{v}\right)−{u}\left({H}−{v}−{l}\:\mathrm{sin}\:\alpha\right)}{\left({H}−{v}−{l}\:\mathrm{sin}\:\alpha\right)\left({H}−{v}\right)}×{H} \\ $$$$\Rightarrow{s}=\frac{\left({H}−{v}\right)\:\mathrm{cos}\:\alpha+{u}\:\mathrm{sin}\:\alpha}{\left({H}−{v}−{l}\:\mathrm{sin}\:\alpha\right)\left({H}−{v}\right)}×{lH} \\ $$$${or} \\ $$$$\Rightarrow{s}=\frac{\left({H}−{y}\right)\:\mathrm{cos}\:\alpha+{x}\:\mathrm{sin}\:\alpha}{\left({H}−{y}−{l}\:\mathrm{sin}\:\alpha\right)\left({H}−{y}\right)}×{lH} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *