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Ray Optics and Optical Instruments Set-4 MCQ Online Practice Test | Class 12 | Quiz Tweek

Ray Optics and Optical Instruments Set-4

Practice Important MCQs for Ray Optics and Optical Instruments — Physics, Class 12.

This chapter is crucial because it forms the backbone of how mirrors and lenses form images, how diffraction limits resolution, and how optical instruments (microscope and telescope) magnify. Board questions often test the lens/magnification formulas and image formation logic, while competitive exams emphasize sign conventions, lens combinations, aberration (chromatic effects), and diffraction-based limits—skills that directly raise your scoring potential.

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Q1. A thin convex lens of focal length $20\ \text{cm}$ forms an image of a $2.0\ \text{cm}$ tall object placed $60\ \text{cm}$ from the lens. Using the lens formula $\displaystyle\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$ (distances measured from the lens), the real image is:

(A)

Image at $40\ \text{cm}$ from the lens; image height $1.33\ \text{cm}$

(B)

Image at $15\ \text{cm}$ from the lens; image height $0.50\ \text{cm}$

(C)

Image at $30\ \text{cm}$ from the lens; image height $1.00\ \text{cm}$

(D)

Image at $20\ \text{cm}$ from the lens; image height $0.67\ \text{cm}$

Q2. Two thin convex lenses L1 and L2, each of focal length $10\ \text{cm}$ , are placed coaxially with their centres $20\ \text{cm}$ apart. An object of height $2.0\ \text{cm}$ is placed $30\ \text{cm}$ to the left of L1. Using $\displaystyle\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$ and taking real objects on the left as positive $u$ , the final image formed by the two-lens system (give position measured from L2, nature, orientation and total magnification) is:

(A)

Final image is $10\ \text{cm}$ to the left of L2 (i.e., between the lenses), virtual, inverted relative to the original, total magnification $m=-1$ , image height $2.0\ \text{cm}$

(B)

Final image is $10\ \text{cm}$ to the right of L2, real, inverted, $m=-1$ , image height $2.0\ \text{cm}$

(C)

Final image is $30\ \text{cm}$ to the right of L2, real, inverted, $m=-0.5$ , image height $1.0\ \text{cm}$

(D)

Final image is $35\ \text{cm}$ to the left of L2, virtual, erect, $m=+0.5$ , image height $1.0\ \text{cm}$

Q3. Two thin lenses L1 (convex) with focal length $+12\ \text{cm}$ and L2 (concave) with focal length $-8\ \text{cm}$ are placed coaxially with separation $16\ \text{cm}$ . An object of height $1.0\ \text{cm}$ is located $24\ \text{cm}$ to the left of L1. Using $\,\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}\,$ for each lens, the final image produced by the combination is:

(A)

A real image $8\ \text{cm}$ to the right of L2, inverted, image height $1.0\ \text{cm}$

(B)

No finite image — emergent rays from L2 are parallel (final image at infinity)

(C)

A virtual image $4\ \text{cm}$ to the left of L1, erect, image height $0.5\ \text{cm}$

(D)

A virtual image $8\ \text{cm}$ to the right of L2, erect, image height $1.0\ \text{cm}$

Q4. An objective of aperture $D$ and focal length $f$ produces an Airy disk on its focal plane whose linear radius is approximately $r\approx 1.22\,\dfrac{\lambda f}{D}$ (wavelength $\lambda$ ). A Barlow lens is inserted to double the effective focal length $(f\to 2f)$ while keeping aperture $D$ unchanged. Which choice correctly states the change (relative to the original objective alone) in (i) angular resolution $\theta$ (Rayleigh criterion), (ii) linear radius $r$ of the Airy disk on the focal plane, and (iii) peak irradiance at the Airy central maximum?

(A)

(i) $\theta$ halves (improves by factor 2); (ii) $r$ doubles; (iii) peak irradiance halves

(B)

(i) $\theta$ unchanged; (ii) $r$ unchanged; (iii) peak irradiance unchanged

(C)

(i) $\theta$ halves; (ii) $r$ increases fourfold; (iii) peak irradiance reduces to one-quarter

(D)

(i) $\theta$ unchanged; (ii) $r$ doubles; (iii) peak irradiance reduces to one-quarter

Q5. A thin symmetric biconvex lens made of glass with refractive index $n_{\text{glass}}=1.50$ has focal length $f_{\text{air}}$ when placed in air ( $n_{\text{air}}=1.00$ ). The lens is now immersed in oil of refractive index $n_{\text{oil}}=1.40$ . Neglect lens thickness and assume surfaces have radii $+R$ and $-R$ . The new focal length $f_{\text{oil}}$ equals:

(A)

$f_{\mathrm{oil}}=\dfrac{n_{\text{oil}}}{n_{\text{air}}}\,f_{\text{air}}=1.4\,f_{\text{air}}$

(B)

$f_{\mathrm{oil}}=7\,f_{\text{air}}$

(C)

$f_{\mathrm{oil}}=\dfrac{n_{\text{glass}}-1}{\,n_{\text{glass}}-n_{\text{oil}}\,}\;f_{\text{air}}=5\,f_{\text{air}}$

(D)

$f_{\mathrm{oil}}=f_{\text{air}}$

Q6. A thin convex lens of focal length $20\ \mathrm{cm}$ forms an image of an object placed $30\ \mathrm{cm}$ from the lens. The image distance from the lens and the nature of the linear magnification (magnitude and orientation) are respectively:

(A)

$12\ \mathrm{cm},\ m=0.4$ (real and inverted)

(B)

$20\ \mathrm{cm},\ m=0.67$ (virtual and upright)

(C)

$60\ \mathrm{cm},\ m=2$ (real and inverted)

(D)

$-60\ \mathrm{cm},\ m=-2$ (virtual and upright)

Q7. Two thin convex lenses $L_1$ and $L_2$ of focal lengths $f_1=10\ \mathrm{cm}$ and $f_2=20\ \mathrm{cm}$ are placed coaxially with their centres $40\ \mathrm{cm}$ apart. An object is placed $15\ \mathrm{cm}$ to the left of $L_1$ . The final image formed by $L_2$ is:

(A)

virtual, located $20\ \mathrm{cm}$ to the left of $L_2$ (i.e. between the lenses); overall magnification $=4$ (inverted)

(B)

real, located $20\ \mathrm{cm}$ to the right of $L_2$ ; overall magnification $=4$ (inverted)

(C)

virtual, located $60\ \mathrm{cm}$ to the left of $L_2$ ; overall magnification $=0.5$ (upright)

(D)

real, located $30\ \mathrm{cm}$ to the right of $L_2$ ; overall magnification $=2$ (inverted)

Q8. A thin symmetric biconvex lens has each radius $R=20\ \mathrm{cm}$ . The refractive index for violet light ( $\lambda_v=450\ \mathrm{nm}$ ) is $n_v=1.532$ and for red light ( $\lambda_r=650\ \mathrm{nm}$ ) is $n_r=1.514$ . An object is at $50\ \mathrm{cm}$ from the lens. Approximately what is the axial separation between the images formed by violet and red light, and which image lies closer to the lens?

(A)

$\approx 0.65\ \mathrm{cm}$ ; red image closer

(B)

$\approx 3.4\ \mathrm{cm}$ ; violet image closer

(C)

$\approx 0.11\ \mathrm{cm}$ ; red image closer

(D)

$\approx 1.7\ \mathrm{cm}$ ; violet image closer

Q9. A compound microscope uses an objective of focal length $f_o=4\ \text{mm}$ and an eyepiece of focal length $f_e=25\ \text{mm}$ . The tube length (distance between objective and eyepiece principal planes) is $L=160\ \text{mm}$ . Take least distance of distinct vision $D=25\ \text{cm}$ . The total magnification for (i) final image at infinity (relaxed eye) and (ii) final image at near point (maximum accommodation) are respectively approximately:

(A)

$160,\ 176$

(B)

$400,\ 440$

(C)

$200,\ 220$

(D)

$320,\ 352$

Q10. An astronomical telescope is in normal adjustment. The objective focal length is $f_o=2.0\ \mathrm{m}$ , the eyepiece focal length is $f_e=25\ \mathrm{mm}$ and the objective diameter is $D=8.0\ \mathrm{cm}$ . Two stars have angular separation $\alpha=2.0\times10^{-5}\ \mathrm{rad}$ . Observations are made at wavelength $\lambda=600\ \mathrm{nm}$ . Which statement is correct?

(A)

Angular magnification $=80$ , separation at objective focal plane $=4.0\times10^{-6}\ \mathrm{m}$ , stars not resolvable

(B)

Angular magnification $=40$ , separation at objective focal plane $=8.0\times10^{-5}\ \mathrm{m}$ , stars resolvable

(C)

Angular magnification $=80$ , separation at objective focal plane $=4.0\times10^{-5}\ \mathrm{m}$ , stars not resolvable

(D)

Angular magnification $=80$ , separation at objective focal plane $=4.0\times10^{-5}\ \mathrm{m}$ , stars resolvable (since $1.22\lambda/D\approx9.15\times10^{-6}\ \mathrm{rad}<\alpha$ )

Q11. A thin convex lens of focal length $15\ \mathrm{cm}$ produces an image of an object placed $20\ \mathrm{cm}$ in front of it. Determine the image distance from the lens and the signed linear magnification.

(A)

$v=60\ \mathrm{cm},\; m=+3$

(B)

$v=60\ \mathrm{cm},\; m=-3$

(C)

$v=12\ \mathrm{cm},\; m=-0.6$

(D)

$v=-60\ \mathrm{cm},\; m=+3$

Q12. Two thin converging lenses L1 and L2 with focal lengths $f_1=15\ \mathrm{cm}$ and $f_2=20\ \mathrm{cm}$ are placed coaxially with their centres $40\ \mathrm{cm}$ apart. An object is placed $30\ \mathrm{cm}$ to the left of L1. Find the position of the final image measured from L2 and state its nature (real/virtual and inverted/upright).

(A)

Final image $20\ \mathrm{cm}$ to the right of L2; real and inverted

(B)

Final image $10\ \mathrm{cm}$ to the left of L2; virtual and inverted

(C)

Final image $30\ \mathrm{cm}$ to the left of L2; real and upright

(D)

Final image $20\ \mathrm{cm}$ to the left of L2; virtual and inverted

Q13. A thin convex lens made of glass with refractive index $n_\ell=1.50$ has focal length $f_{\text{air}}=20\ \mathrm{cm}$ when used in air. The same lens is immersed in a liquid of refractive index $n_m=1.33$ . Calculate its new focal length in the liquid.

(A)

$f_{\text{liquid}}\approx 78.22\ \mathrm{cm}$

(B)

$f_{\text{liquid}}\approx 13.56\ \mathrm{cm}$

(C)

$f_{\text{liquid}}\approx 33.33\ \mathrm{cm}$

(D)

$f_{\text{liquid}}\approx 22.67\ \mathrm{cm}$

Q14. An achromatic doublet is made by placing in contact a crown-glass lens (Abbe number $V_1=58$ ) and a flint-glass lens ( $V_2=40$ ). The combined focal length of the doublet is $F=20\ \mathrm{cm}$ . Using the achromatic condition, determine the focal lengths $f_1$ (crown) and $f_2$ (flint) including their signs.

(A)

$f_1\approx +9.00\ \mathrm{cm},\; f_2\approx +6.21\ \mathrm{cm}$

(B)

$f_1\approx -6.21\ \mathrm{cm},\; f_2\approx +9.00\ \mathrm{cm}$

(C)

$f_1\approx +6.21\ \mathrm{cm},\; f_2\approx -9.00\ \mathrm{cm}$

(D)

$f_1\approx +12.42\ \mathrm{cm},\; f_2\approx -24.84\ \mathrm{cm}$

Q15. A plane mirror is placed immediately behind a glass slab of thickness $t=4.0\ \mathrm{cm}$ . An object is placed in front of the slab; when the slab is inserted (so light passes through the slab, reflects off the mirror and passes back through the slab), the image formed by the mirror is observed to shift away from the object by $3.2\ \mathrm{cm}$ compared to the case without the slab. Determine the refractive index $n$ of the slab (assume paraxial approximation).

(A)

$n=1.32$

(B)

$n=1.40$

(C)

$n=1.20$

(D)

$n=2.00$

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