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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Physics
Chapter
Ray Optics

JEE Advanced 2016 · Paper 2 · Physics · Ray Optics

A small object is placed 50 cm to the left of thin convex lens of...

A small object is placed 50 cm to the left of thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle θ=30o\theta = {30}^{o} to the axis of the lens, as shown in the figure. If the origin of the coordinate 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 are:
A small object is placed 50 cm to the left of thin convex lens of focal length 30 cm. A convex spherical mirror of

Options

  1. A(25,253)\left(25, 25\sqrt{3}\right)
  2. B(1253,253)\left(\dfrac{125}{3}, \dfrac{25}{\sqrt{3}}\right)
  3. C(50−253,25)\left(50 - 25\sqrt{3}, 25\right)
  4. D(0,0)\left(0, 0\right)

Correct answer

A. (25,253)\left(25, 25\sqrt{3}\right)

Step-by-step solution

A small object is placed 50 cm to the left of thin convex lens of focal length 30 cm. A convex spherical mirror of
For lens V=(−50)(30)−50+30=75V = \dfrac{( - 50 ) ( 30 )}{- 50 + 30} = 75
For mirror V=(2532)(50)2532−50=−5034−3V = \dfrac{\left(\dfrac{25\sqrt{3}}{2}\right)\left(50\right)}{\dfrac{25\sqrt{3}}{2} - 50} = \dfrac{- 50\sqrt{3}}{4 - \sqrt{3}}
m=vu=h2h1⇒h2=−(−5034−32532).252m = \dfrac{v}{u} = \dfrac{{h}_{2}}{{h}_{1}} ⇒ {h}_{2} = - \left(\dfrac{\dfrac{- 50\sqrt{3}}{4 - \sqrt{3}}}{\dfrac{25\sqrt{3}}{2}}\right) . \dfrac{25}{2}
h2=+504−3{h}_{2} = \dfrac{+ 50}{4 - \sqrt{3}}
The x coordinate of the images =50−vcos30+h2cos⁡60≈25= 50 - vcos 30 + {h}_{2}\cos 60 \approx25
The y coordinate of the images =vsin30+h2sin⁡60≈253= vsin 30 + {h}_{2}\sin 60 \approx 25\sqrt{3}
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