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Step-by-Step Solution
Step 1: Identify the Type of Lenses and Given Data
• We have two thin lenses:
(i) a plano concave lens (with one surface flat and the other concave), and
(ii) a plano convex lens (with one surface flat and the other convex).
• Both lenses have the same radius of curvature R = 30\,\text{cm} .
• The refractive index of the material for both lenses is n = 1.75 .
• The object is placed at infinity.
• The distance between the two lenses is 40\,\text{cm} .
• We want to find the final image distance x from the concave lens.
Step 2: Compute Focal Length of the Plano Concave Lens
The lens maker’s formula for a thin lens in air is
\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right).
For a plano concave lens, R_1 = \infty (flat surface) and R_2 = +30\,\text{cm} (the sign is positive if we consider the convention that the center of curvature is on the outgoing side for a concave surface, but effectively it will lead to a negative focal length). Thus,
\frac{1}{f_{\text{concave}}} = (1.75 - 1)\left(\frac{1}{\infty} - \frac{1}{+30}\right)
= 0.75\left(0 - \frac{1}{30}\right)
= -\frac{0.75}{30}.
Therefore,
f_{\text{concave}}
= -\frac{30}{0.75}
= -40\,\text{cm}.
The negative sign indicates that the plano concave lens has a negative focal length (diverging lens).
Step 3: Compute Focal Length of the Plano Convex Lens
For the plano convex lens, R_1 = 30\,\text{cm} and R_2 = \infty . Using the lens maker's formula again,
\frac{1}{f_{\text{convex}}} = (1.75 - 1)\left(\frac{1}{30} - \frac{1}{\infty}\right)
= 0.75 \left(\frac{1}{30} - 0\right)
= \frac{0.75}{30}.
Hence,
f_{\text{convex}}
= \frac{30}{0.75}
= 40\,\text{cm}.
The positive sign indicates this lens is converging.
Step 4: Find the Image Formed by the Plano Concave Lens
The object is at infinity for the concave lens. Using the lens formula
\frac{1}{v} - \frac{1}{u} = \frac{1}{f},
with u = \infty and f_{\text{concave}} = -40\,\text{cm} , we get:
\frac{1}{v_1} - \frac{1}{\infty} = \frac{1}{-40}
\quad\Longrightarrow\quad
\frac{1}{v_1} = -\frac{1}{40}.
Thus,
v_1 = -40\,\text{cm}.
The image is formed 40\,\text{cm} on the same side of the lens as the object (virtual image for the concave lens).
Step 5: Use the Image from the First Lens as the Object for the Second Lens
The distance between the two lenses is 40\,\text{cm} . Since the concave lens produces a virtual image at v_1 = -40\,\text{cm} , the object distance for the convex lens ( u_2 ) is measured from the convex lens. Let d be the separation between the two lenses. We have:
• The virtual image is 40\,\text{cm} behind the concave lens.
• The convex lens is 40\,\text{cm} away from the concave lens.
Therefore, the location of that virtual image from the convex lens is u_2 = -(d + |v_1|) = -(40 + 40)\,\text{cm} = -80\,\text{cm}.
In other words, for the convex lens:
u_2 = -80\,\text{cm},
\quad
f_{\text{convex}} = 40\,\text{cm}.
Step 6: Determine the Final Image Distance from the Concave Lens
Again, applying the lens formula for the convex lens:
\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_{\text{convex}}}.
Here,
\frac{1}{v_2} - \frac{1}{-80} = \frac{1}{40}.
Substituting the values,
\frac{1}{v_2} + \frac{1}{80} = \frac{1}{40}.
Hence,
\frac{1}{v_2} = \frac{1}{40} - \frac{1}{80}
= \frac{2 - 1}{80}
= \frac{1}{80}.
Therefore,
v_2 = 80\,\text{cm}.
Note that this v_2 is the distance from the convex lens. We want x , the distance of this final image from the concave lens. We have to add the separation between the lenses ( 40\,\text{cm} ) back:
x = d + v_2 = 40 + 80 = 120\,\text{cm}.
Final Answer
The final image is formed at a distance of
x = 120\,\text{cm}
from the concave lens.
