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Step 1: Identify the Refractive Index of the Lens Material
Given that the wavelength of light inside the lens, \lambda_{\text{med}} , is \tfrac{3}{4} times the wavelength in free space, \lambda_{\text{vacuum}} , the refractive index \mu of the lens material can be determined as follows:
\mu \;=\; \dfrac{\lambda_{\text{vacuum}}}{\lambda_{\text{med}}}
\;=\; \dfrac{1}{\tfrac{3}{4}}
\;=\; \dfrac{4}{3}.
Step 2: Use Magnification to Find the Object Distance
We know the image distance (real image) behind the lens is v = +10\,\text{m} . The image is one-fourth the size of the object, so the linear magnification m is -\tfrac{1}{4} (negative sign implies a real, inverted image):
m \;=\; \dfrac{v}{u} \;=\; -\dfrac{1}{4}.
From this, solve for the object distance u :
u \;=\; \dfrac{v}{m}
\;=\; \dfrac{10}{-\tfrac{1}{4}}
\;=\; -40\,\text{m}.
(A negative sign indicates the object is placed on the same side of the lens as the incident light.)
Step 3: Apply the Lens Formula to Determine the Focal Length
The lens formula is:
\dfrac{1}{f} \;=\; \dfrac{1}{v} \;-\; \dfrac{1}{u}.
Substitute v = 10\,\text{m} and u = -40\,\text{m} :
\dfrac{1}{f}
\;=\; \dfrac{1}{10} \;-\; \left(\dfrac{1}{-40}\right)
\;=\; \dfrac{1}{10} \;+\; \dfrac{1}{40}
\;=\; \dfrac{4}{40} \;+\; \dfrac{1}{40}
\;=\; \dfrac{5}{40}
\;=\; \dfrac{1}{8}.
Thus, f = 8\,\text{m}.
Step 4: Relate Focal Length and Radius of Curvature for a Plano-Convex Lens
A plano-convex lens has one flat surface (radius of curvature R = \infty ) and one curved surface of radius R . Its focal length is given by:
\dfrac{1}{f} \;=\; (\mu - 1)\,\dfrac{1}{R}.
Rearranging for R :
R = f\,(\mu - 1)
= 8 \left(\dfrac{4}{3} - 1\right)
= 8 \times \tfrac{1}{3}
= \dfrac{8}{3}.
Step 5: Determine the Value of \alpha
The problem states that the curved surface radius is given by \tfrac{\alpha}{3} . We have found:
R \;=\; \dfrac{8}{3}.
Thus,
\dfrac{\alpha}{3} \;=\; \dfrac{8}{3}
\quad \Longrightarrow \quad
\alpha \;=\; 8.
