Look at the ray diagram shown. What will be t | vedclass

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(D) $1$. For the $1^{st}$ lens,the incident ray is refracted such that it becomes parallel to the principal axis. This means the incident ray is direc

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$+5 	ext{ cm}$ and $+5 	ext{ cm}$

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(D) $1$. For the $1^{st}$ lens,the incident ray is refracted such that it becomes parallel to the principal axis. This means the incident ray is directed towards the focal point of the $1^{st}$ lens. From the diagram,the distance of this focal point from the $1^{st}$ lens is $5 	ext{ cm}$. Since the lens is convex,its focal length is $f_{1} = +5 	ext{ cm}$.$2$. For the $2^{nd}$ lens,the ray parallel to the principal axis strikes the lens and is refracted to pass through its focal point. The emergent ray makes the same angle $	heta$ with the principal axis as the incident ray. From the geometry of the diagram,$	an 	heta = \frac{h}{5} = \frac{h}{f_{2}}$,where $h$ is the height of the ray from the principal axis. Thus,the focal length of the $2^{nd}$ lens is $f_{2} = 5 	ext{ cm}$. Since it is a convex lens,$f_{2} = +5 	ext{ cm}$.

Look at the ray diagram shown. What will be the focal length of the $1^{st}$ and the $2^{nd}$ lens,if the incident light ray passes without any net deviation?

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