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(B) For the first lens:
Using the lens formula $\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1}$, where $f_1 = 2 \ m$ and $u_1 = -4 \ m$:
$\frac{1}{2}

Vedclass · Accepted Answer

$7.5$

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(B) For the first lens:
Using the lens formula $\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1}$, where $f_1 = 2 \ m$ and $u_1 = -4 \ m$:
$\frac{1}{2} = \frac{1}{v_1} - \frac{1}{-4} \Rightarrow \frac{1}{v_1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \Rightarrow v_1 = 4 \ m$.
This image is formed $4 \ m$ to the right of the first lens.
For the second lens:
The distance between the lenses is $3 \ m$. The image formed by the first lens acts as an object for the second lens.
Since the image is $4 \ m$ to the right of the first lens and the second lens is $3 \ m$ to the right of the first, the image is $1 \ m$ to the right of the second lens.
Thus, the object distance for the second lens is $u_2 = +1 \ m$ (virtual object).
Using the lens formula $\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}$, where $f_2 = 1 \ m$ and $u_2 = +1 \ m$:
$\frac{1}{1} = \frac{1}{v_2} - \frac{1}{1} \Rightarrow \frac{1}{v_2} = 1 + 1 = 2 \Rightarrow v_2 = 0.5 \ m$.
The final image is $0.5 \ m$ to the right of the second lens.
The total distance from the source is $4 \ m$ (to first lens) $+ 3 \ m$ (between lenses) $+ 0.5 \ m$ (from second lens) $= 7.5 \ m$.

An object is located $4 \ m$ from the first of two thin converging lenses of focal lengths $2 \ m$ and $1 \ m,$ respectively. The lenses are separated by $3 \ m$. The final image formed by the second lens is located from the source at a distance of (in $m$):

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