Ray Optics Question 1

Question 1 - 24 January - Shift 1

As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is $30 cm$ and refraction index of the material for both the lenses is 1.75 . Both the lenses are placed at distance of 40 $cm$ from each other. Due to the combination, the image of the object is formed at distance $x=$ $cm$, from concave lens.

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Answer: (120)

Solution:

Formula: Lens Maker’s Formula

$\frac{1}{f_1}=(1.75-1)(-\frac{1}{30})$

$\Rightarrow f_1=-40 cm$

$\frac{1}{f_2}=(1.75-1)(\frac{1}{30}) \Rightarrow f_2=40 cm$

Image from $L_1$ will be virtual and on the left of $L_1$ at focal length $40 cm$. So the object for $L_2$ will be $80 cm$ from $L_2$ which is $2 f$. Final image is formed at $80 cm$ from $L_2$ on the right.

So $x=120$