Ray Optics and Optical Instruments Set-3

Q1. A thin equiconvex lens has both surfaces of radius $R=20\ \text{cm}$ and is made of glass with refractive index $n=1.5$. Using the lens‑maker formula $\displaystyle \frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ and $R_1=+R,\ R_2=-R$, the focal length $f$ of the lens is:

Q2. For a thin convex lens the image distance $v$ and object distance $u$ (measured from the lens) satisfy $v=\dfrac{u f}{\,u-f\,}$. If the graph of $v$ versus $u$ (for $u>f$) has a vertical asymptote at $u=f$ and passes through $(2f,2f)$, the image distance $v$ when $u=1.5f$ equals:

Q3. A thin converging lens made of glass ($n_{\ell}=1.6$) has focal length $f_{\rm air}=20\ \text{cm}$ in air. It is immersed in a liquid of refractive index $n_{\rm liq}=1.5$. Using the relation for a lens in a medium $\,\frac{1}{f_{\rm med}}=\Big(\frac{n_{\ell}}{n_{\rm med}}-1\Big)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\,$ and eliminating the curvature terms with the known $f_{\rm air}$, the new focal length $f_{\rm liq}$ (in cm) is approximately:

Q4. Two thin lenses in contact have focal lengths $f_1=+20\ \text{cm}$ (convex) and $f_2=-30\ \text{cm}$ (concave). The equivalent focal length $F$ of the combination is given by $\,\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_2}\,$. The value of $F$ (in cm) is:

Q5. Assertion (A): For a glass–air interface the critical angle for blue light is smaller than that for red light.
Reason (R): This follows because the refractive index of typical optical glass is larger for blue than for red (normal dispersion).

Q6. A thin convex lens has focal lengths $f_{\rm red}=20.0\ \text{cm}$ (red) and $f_{\rm blue}=19.6\ \text{cm}$ (blue). An object at $u=40.0\ \text{cm}$ is sharply imaged on a screen for red light. When illuminated with blue light without moving lens or screen, the blue image forms at a different position. Approximately how far and in which direction is the blue image relative to the screen set for red light?

Q7. In a compound microscope the objective has focal length $f_o=2.0\ \text{cm}$ and the eyepiece has focal length $f_e=5.0\ \text{cm}$. The tube length is $L=16.0\ \text{cm}$. For the relaxed‑eye case (final image at infinity) and taking the near point $D=25\ \text{cm}$, the approximate total angular magnification $M\simeq\left(\dfrac{L}{f_o}\right)\left(\dfrac{D}{f_e}\right)$. The value of $M$ is:

Q8. Assertion (A): Reducing the aperture (stopping down) of a lens reduces spherical aberration but does not eliminate chromatic aberration.
Reason (R): Spherical aberration arises from marginal rays focusing at different points and stopping the aperture removes marginal rays; chromatic aberration is due to wavelength‑dependent refractive index and cannot be removed by reducing aperture alone.

Q9. In an experiment with an equilateral prism ($A=60^\circ$) the measured minimum deviation is $D_{\min}=40^\circ$. Using $\,n=\dfrac{\sin\big(\tfrac{A+D_{\min}}{2}\big)}{\sin\big(\tfrac{A}{2}\big)}\,$, the refractive index $n$ of the prism material is approximately:

Q10. For a thin lens the image distance $v$ as a function of object distance $u$ satisfies $v=\dfrac{u f}{u-f}$ and $\dfrac{dv}{du}=-\dfrac{f^2}{(u-f)^2}$. For which of the following object distances (measured in multiples of $f$) does a small change in $u$ produce the largest magnitude change in $v$?