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A small object is placed 50cm to the left of a thin convex lens of focal length 30cm. A convex spherical mirror of radius of curvature 100cm is placed to the right of the lens at a distance of 50cm. The lens at a distance of 50cm. The mirror is titled such that the axis of the mirror is at an angle $\theta = {30^o}$to the axis of the lens, as shown in the figure. If the origin of the coordinates system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x,y) at which the image is formed is:
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Last updated date: 20th Jun 2024
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Answer
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Hint:First find the image formed by the lens, now image of the lens will serve as a object for the mirror but principle axis of the of the mirror is different from the lens so you will have to find the position of the image by lens or object of mirror with respect to the principle axis of the mirror once you know the position accurately then use mirror formula to find the coordinates of image.

Complete step by step answer:
We will find the image formed by the lens first,for lens object distance is 50cm, focal length is 30cm,i.e; u=-50cm and f=30cm.Using lens formula we have,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
$ \Rightarrow \dfrac{1}{{30}} = \dfrac{1}{v} - \dfrac{1}{{ - 50}}$
$ \Rightarrow v = \dfrac{{50 \times 30}}{{50 - 30}}\\
\therefore v = 25cm$

So along the principle axis of lens distance of object from the pole of the mirror will be
$d = 75 - 50 = 25cm$
Now angle of inclination of principal axis of mirror with the principle of lens is 30degrees
So object distance for mirror will be
${u_1} = 25\cos {30^o} \\
\Rightarrow{u_1}= 12.5\sqrt 3 $
Radius of curvature of the mirror is given taking half of the radius of curvature. We can find the value of focal length of the mirror.
So $f = \dfrac{R}{2}\\
\Rightarrow f = \dfrac{{100}}{2} \\
\therefore f = 50$

Using mirror formula we have
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
$ \Rightarrow \dfrac{1}{{50}} = \dfrac{1}{v} + \dfrac{1}{{12.5\sqrt 3 }}$
$\Rightarrow v = \dfrac{{12.5\sqrt 3 \times 50}}{{12.5\sqrt 3 - 50}}\\
\therefore v= \dfrac{{ - 50\sqrt 3 }}{{4 - \sqrt 3 }}$
Also height of object for mirror will be,
${h_O} = 25\sin {30^ \circ }
\Rightarrow{h_O}= \dfrac{{25}}{2}cm$

Using magnification of mirror height of image will be
${h_I} = m{h_O}\\
\Rightarrow{h_I}= \dfrac{{ - v}}{u}{h_O}$
$\Rightarrow {h_I} = - \dfrac{{\dfrac{{ - 50\sqrt 3 }}{{4 - \sqrt 3 }}}}{{12.5\sqrt 3 }} \times \dfrac{{25}}{2}\\
\therefore {h_I} = \dfrac{{50}}{{4 - \sqrt 3 }}$
The x-coordinates of the image will be $50 - v\cos 30 + {h_I}\cos 60 \approx 25$.
The y-coordinate of the image will be $v\sin 30 + {h_I}\sin 60 \approx 25\sqrt 3 $.

Hence coordinates of the image will be $(25,25\sqrt 3 )$.

Note: Image formed by the lens will serve as a object for the mirror but principal axis of the of the mirror is different from the lens so you will have to find the position of the image by lens or object of mirror with respect to the principal axis of the mirror once you know the position accurately then use mirror formula to find the coordinates of image.