Question

we are given a concave lens of focal length 15 CM draw ray diagram to show the nature size and position of the image formed when the object is kept at a distance of 30cm​

Answer 1

$\setlength{\unitlength}{1 cm}\begin{picture}{0}{\thicklines}\put(2,2){\line(0,2){2}}\put(2,2){\line(2,0){3}}\put(5,2){\line(0,2){2}}\put(3,2){\line(1,1){2}}\put(3,2){\line(-1,2){1}}\put(5.2,3){\large{h}}\put(1.6,3){\large{h}}\put(2.3,1.5){\large{x}}\put(3.4,1.5){\large{100 - x}}\put(2.3,2.1){60^{\circ}}}\put(0.6,-0){30^{\circ}}}\put(1.1,1.9){D}\put(-2.3,-0.3){A}\put(-1,-0.5){B}\put(1.1,-0.3){C}\put(-2.2,1.9){E}\end{picture}$

$\LARGE\bold{Given :}$

Bottom distance = 100 m.

Angle of Elevation$\bf{\angle_{1}}$

Angle of Elevation$\bf{\angle_{2}}$

$\LARGE\bold{To\: find :}$

Height of the two poles.

Distance of the points from the feet of the poles.

$\LARGE\bold{Solution :}$

Let the height of both the poles will be h m.

Let the distance from point A and B be x m.

Hence according to the Question , the distance from point B will be (100 - x) m.

$\LARGE\bold{Height \:of\: the\: tower :)}$

To find the height of pole (in terms of h) with respect to angle 60°.

Using tan θ and substituting the values in it, we get :

$\begin{gathered}:\implies \bf{tan\:\theta = \dfrac{P}{B}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{tan\:60^{\circ} = \dfrac{h}{x}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{\sqrt{3} = \dfrac{h}{x}}\quad[\because \bf{tan\:60^{\circ} = \sqrt{3}}] \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{x = \dfrac{h}{\sqrt{3}}} \\ \\ \\\end{gathered}$

$\begin{gathered}\boxed{\therefore \bf{x = \dfrac{h}{\sqrt{3}}}} \quad \quad Eq..(i) \ \\ \\ \\\end{gathered}$

Hence the distance between base of A and B (in terms of h) is √3/h

Now , by using the tan θ and substituting the values in it, we get :

$\begin{gathered}:\implies \bf{tan\:\theta = \dfrac{P}{B}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{tan\:30^{\circ} = \dfrac{h}{100 - x}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{\dfrac{1}{\sqrt{3}} = \dfrac{h}{100 - x}}\quad[\because \bf{tan\:30^{\circ} = \dfrac{1}{\sqrt{3}}}] \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{\dfrac{100 - x}{\sqrt{3}} = h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{\dfrac{100 - x}{\sqrt{3}} = h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100 - x = h\sqrt{3}} \\ \\ \\\end{gathered}$

Now , by substituting the value of x from equation (i) , we get :

$\begin{gathered}:\implies \bf{100 - \dfrac{h}{\sqrt{3}} = h\sqrt{3}} \\ \\ \\\end{gathered}$

$@\begin{gathered}:\implies \bf{\dfrac{100\sqrt{3} - h}{\sqrt{3}} = h\sqrt{3}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100\sqrt{3} - h = h\sqrt{3} \times \sqrt{3}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100\sqrt{3} - h = 3h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100\sqrt{3} = h + 3h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100\sqrt{3} = 4h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{\dfrac{100\sqrt{3}}{4} = h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{25\sqrt{3} = h} \\ \\ \\\end{gathered}$

$\begin{gathered}\boxed{\therefore \bf{h = 25\sqrt{3}}} \\ \\ \\\end{gathered}$

Hence the Height of two towers is 25√3 m.

$\LARGE\bold{Distance\: from\: the \:points\: :}$

$\LARGE\bold{Distance \:between \:A\: and \:B\: :}$

Since, we have taken the base distance as x and we know the value of x in terms of h i.e,

$\begin{gathered}:\implies \bf{x = \dfrac{h}{\sqrt{3}}} \\ \\ \\\end{gathered}$

Now, putting the value of h in the above equation , we get :

$\begin{gathered}:\implies \bf{x = \dfrac{25\sqrt{3}}{\sqrt{3}}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{x = 25 m} \\ \\ \\\end{gathered}$

$\begin{gathered}\boxed{\therefore \bf{x = 25\:m}} \\ \\ \\\end{gathered}$

Hence, the base distance from A to B is 25.

Distance between B and C :

We know that the distance between B and C is (100 - x) m.

So by putting the value of x in it , we get :

$\begin{gathered}:\implies \bf{100 - x} \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100 - 25} \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{75} \\ \\\end{gathered}$

$\begin{gathered}\boxed{\therefore \bf{75\:m}} \\ \\ \\\end{gathered}$

Hence the distance between B and C is 75 m.

Answer 2

Answer:

Let object distance be C. Then the image distance will also be 'C' . Because in case of concave mirror, If the object is kept at C , the image also forms at 'C'.

Given:

$\sf{f=-15cm}$

$\sf{\therefore {C=-30cm}}$

$\sf{u=-30cm}$

$\sf{Mirror \:= \:Concave \:mirror}$

Proof:

$\sf{{\dfrac{1}{f}}={\dfrac{1}{v}}+{\dfrac{1}{u}}}$

$\sf{-{\dfrac{1}{15}}={\dfrac{1}{v}}-{\dfrac{1}{30}}}$

$\sf{{\dfrac{1}{v}}=-{\dfrac{1}{15}}+{\dfrac{1}{30}}}$

$\sf{{\dfrac{1}{v}}=-{\dfrac{1}{30}}}$

$\sf{v=-30cm}$

The object formed thus will be real and inverted and of the same size as the object.