Ray Optics and Optical Instruments Set-5

Q1. An object is placed $20\ \text{cm}$ to the left of a thin converging lens and a real image is formed $30\ \text{cm}$ to the right of the lens. Using the thin-lens formula $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$ (with $u$ and $v$ as positive magnitudes), the focal length $f$ of the lens is:

Q2. Two thin convex lenses $L_1$ and $L_2$ with focal lengths $f_1=20\ \text{cm}$ and $f_2=30\ \text{cm}$ are coaxial and separated by $40\ \text{cm}$. An object of height $1.0\ \text{cm}$ is placed $60\ \text{cm}$ to the left of $L_1$. Using $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$ and magnification $m=-\dfrac{v}{u}$, the final image formed by the two-lens system is:

Q3. A thin converging lens of focal length $f=15\ \text{cm}$ is fixed with a screen placed $80\ \text{cm}$ to its right. Show that there are two object positions on the left of the lens for which a sharp image appears on the screen. Find the two object distances and the corresponding linear magnifications (use $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$ and $m=-\dfrac{v}{u}$).

Q4. A thin converging lens of focal length $f=20\ \text{cm}$ forms a real image exactly at $2f$ when an object is at $2f$ (so initial image at $40\ \text{cm}$). A plane-parallel glass slab of thickness $t=2\ \text{cm}$ and refractive index $\mu=1.5$ is inserted between the object and the lens (slab faces perpendicular to the axis). Using the apparent shift by a slab $\Delta=t\big(1-\tfrac{1}{\mu}\big)$ and the thin-lens formula $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$, the new image position measured from the lens and its nature are:

Q5. Assertion (A): In an astronomical telescope adjusted in normal adjustment the angular magnification is $M=-\dfrac{f_o}{f_e}$, where $f_o$ and $f_e$ are the focal lengths of the objective and eyepiece respectively.
Reason (R): In normal adjustment the separation between objective and eyepiece is $f_o+f_e$, so the image formed by the objective lies at the focal plane of the eyepiece and, using small-angle approximations, one obtains $M=-\dfrac{f_o}{f_e}$.

Q6. A thin convex lens has focal length $f=10\ \text{cm}$. An object of height $2.0\ \text{cm}$ is placed $15\ \text{cm}$ from the lens. Using the thin-lens relation $v=\dfrac{fu}{u-f}$ and magnification $m=\dfrac{v}{u}$ (sign of $m$ indicates inversion), determine the image distance and the image height.

Q7. Two thin convex lenses L₁ and L₂ with focal lengths $f_1=10\ \text{cm}$ and $f_2=20\ \text{cm}$ are coaxial and separated by $35\ \text{cm}$. An object is placed $15\ \text{cm}$ to the left of L₁. By applying the thin-lens formula successively ($v=\dfrac{fu}{u-f}$ for each lens), determine the final image position measured from L₂ and state whether it is real or virtual.

Q8. Assertion (A): For a simple magnifying glass used with the eye relaxed (final image at infinity), increasing the focal length of the eyepiece increases its angular magnification. Reason (R): For a relaxed eye the angular magnification of a simple magnifier (using $D=25\ \text{cm}$ as the least distance of distinct vision) is $M=\dfrac{25}{f}$ (with $f$ in cm).

Q9. A concave mirror of focal length $f=15\ \text{cm}$ has an object placed $40\ \text{cm}$ in front of it. A glass slab of thickness $t=10\ \text{cm}$ and refractive index $n=1.5$ is inserted between the object and the mirror (faces parallel to mirror). Treat the slab effect as an effective reduction of the object distance by $\Delta=2t\!\left(1-\dfrac{1}{n}\right)$ (rays pass through slab twice) and use the mirror formula $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$ to find the new image distance from the mirror and the nature of the image.

Q10. A compound microscope has an objective focal length $f_o=4\ \text{mm}$ and an eyepiece focal length $f_e=25\ \text{mm}$. The tube length (distance between the objective's image plane and the eyepiece's object plane) is $L=160\ \text{mm}$. If the intermediate image formed by the objective lies at $v_o=L$ and the final image is at infinity (eye relaxed), estimate the total magnifying power using $M_{\rm total}=m_o\times M_e$ with $m_o\approx\dfrac{v_o}{u_o}$, $u_o=\dfrac{v_o f_o}{v_o-f_o}$ and $M_e\approx\dfrac{25\ \text{cm}}{f_e}$ (use consistent units).

Q11. A thin convex lens of focal length $20\ \text{cm}$ produces a real image of an object placed $30\ \text{cm}$ from the lens. Using the lens formula $\displaystyle\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ (take $u>0$ for a real object), find the image distance $v$ and the image height if the object height is $2\ \text{cm}$ (give magnitude of image height).

Q12. Two thin coaxial lenses L1 and L2 are separated by $25\ \text{cm}$. L1 (left) is converging with $f_1=+15\ \text{cm}$ and L2 (right) is diverging with $f_2=-10\ \text{cm}$. An object is placed $30\ \text{cm}$ to the left of L1. Using $\displaystyle\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ for each lens, determine the final image distance measured from L2 (sign indicates side) and the overall linear magnification $m$.

Q13. A symmetric biconvex glass lens has refractive index $n_g=1.50$ and both radii $R_1=+20\ \text{cm},\ R_2=-20\ \text{cm}$. Using the generalized lens-maker formula $\displaystyle\frac{1}{f}=\Big(\frac{n_{\text{lens}}}{n_{\text{med}}}-1\Big)\!\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)$, compute (i) its focal length in air ($n_{\text{med}}=1.00$) and (ii) its focal length when immersed in a liquid of refractive index $n_\ell=1.33$.

Q14. Assertion (A): A biconvex glass lens (refractive index $1.50$) immersed in a medium whose refractive index exceeds $1.50$ will behave as a diverging lens.
Reason (R): The nature (converging or diverging) of a lens is determined only by the signs of its surface curvatures and is independent of the refractive indices of the lens and the surrounding medium.

Q15. A Keplerian telescope has an objective of focal length $f_o=1200\ \text{mm}$ and diameter $D=80\ \text{mm}$ and an eyepiece of focal length $f_e=20\ \text{mm}$. For light of wavelength $\lambda=550\ \text{nm}$ and an observer whose pupil diameter is $D_{\text{eye}}=6\ \text{mm}$, find the minimum angular separation on the sky that can be resolved when the telescope is in normal adjustment. Use Rayleigh criterion $\theta=1.22\lambda/D$ and note the effective minimum on sky is $\max\!\big(1.22\lambda/D,\ \theta_{\text{eye}}/M\big)$ with $M=f_o/f_e$ and $\theta_{\text{eye}}=1.22\lambda/D_{\text{eye}}$. Give the result in arcseconds (″).