Ray Optics and Optical Instruments MCQs with Answers – Part 4 (Class 12 Physics)

Explanation: The equivalent power of a lens system in contact is related to equivalent focal length by \(P_{\text{eq}}=\frac{1}{f_{\text{eq}}}\). A negative power means that the system is diverging. Here, \[ P_{\text{eq}}=-1.5\,\text{D} \] Therefore, \[ f_{\text{eq}}=\frac{1}{P_{\text{eq}}} \] \[ f_{\text{eq}}=\frac{1}{-1.5}\,m \] \[ f_{\text{eq}}=-\frac{2}{3}\,m \] \( \textbf{Final answer:} \) the combination is diverging with focal length \(-\frac{2}{3}\,m\).

Explanation: \( \textbf{Equivalent power:} \) For thin lenses in contact, powers add algebraically. \[ P_{\text{eq}}=P_1+P_2 \] \[ P_{\text{eq}}=+4\,\text{D}+(-1\,\text{D}) \] \[ P_{\text{eq}}=+3\,\text{D} \] \( \textbf{Equivalent focal length:} \) \[ f=\frac{1}{P_{\text{eq}}}=\frac{1}{3}\,m \] \[ f=\frac{100}{3}\,cm \] \( \textbf{Object distance:} \) \[ u=-60\,cm \] \( \textbf{Lens formula:} \) \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] \( \textbf{Substitution:} \) \[ \frac{3}{100}=\frac{1}{v}-\frac{1}{-60} \] \[ \frac{3}{100}=\frac{1}{v}+\frac{1}{60} \] \[ \frac{1}{v}=\frac{3}{100}-\frac{1}{60} \] \[ \frac{1}{v}=\frac{9-5}{300}=\frac{4}{300} \] \[ \frac{1}{v}=\frac{1}{75} \] \[ v=+75\,cm \] The positive image distance shows that the real image forms on the refracted side of the lens combination. \( \textbf{Final answer:} \) \(+75\,cm\).

Explanation: \( \textbf{Equivalent power first:} \) The two lenses are in contact, so their powers add: \[ P_{\text{eq}}=P_1+P_2 \] \[ P_{\text{eq}}=+5\,\text{D}+(-2\,\text{D})=+3\,\text{D} \] Therefore the equivalent focal length is \[ f=\frac{1}{P_{\text{eq}}}=\frac{1}{3}\,m \] \[ f=\frac{100}{3}\,cm \] The object is real and in front of the lens combination: \[ u=-50\,cm \] Use the thin lens formula: \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] \[ \frac{3}{100}=\frac{1}{v}-\frac{1}{-50} \] \[ \frac{3}{100}=\frac{1}{v}+\frac{1}{50} \] \[ \frac{1}{v}=\frac{3}{100}-\frac{2}{100}=\frac{1}{100} \] \[ v=+100\,cm \] \( \textbf{Final answer:} \) \(+100\,cm\), so the image is formed on the far side of the combination.

Explanation: \( \textbf{Given powers:} \) \[ P_1=+2\,\text{D},\quad P_2=+3\,\text{D},\quad P_3=-1\,\text{D} \] For thin lenses in contact, \[ P_{\text{eq}}=P_1+P_2+P_3 \] Substitute the values: \[ P_{\text{eq}}=2+3-1 \] \[ P_{\text{eq}}=+4\,\text{D} \] Equivalent focal length is \[ f_{\text{eq}}=\frac{1}{P_{\text{eq}}} \] \[ f_{\text{eq}}=\frac{1}{4}\,m \] \[ f_{\text{eq}}=+0.25\,m \] The positive sign shows that the combination is converging. \( \textbf{Final answer:} \) \(+0.25\,m\).

Explanation: For thin lenses in contact, the directly additive quantity is power. The relation is \(P_{\text{eq}}=P_1+P_2+\cdots\), not \(f_{\text{eq}}=f_1+f_2+\cdots\). If \(f_1=+20\,cm=+0.20\,m\), then \(P_1=+5\,\text{D}\). If \(f_2=-20\,cm=-0.20\,m\), then \(P_2=-5\,\text{D}\). The equivalent power is \(0\,\text{D}\), so the ideal equivalent focal length is infinite, not zero. Zero focal length would mean infinitely large power, which is the opposite of \(P_{\text{eq}}=0\,\text{D}\).

Explanation: Row P is acceptable because \(+4-1=+3\,\text{D}\), giving a converging combination. Row Q is acceptable because \(+2-5=-3\,\text{D}\), giving a diverging combination. Row R is acceptable because equal and opposite powers cancel, giving \(0\,\text{D}\) in the ideal contact model. Row S is not acceptable because \(-2-4=-6\,\text{D}\), so the combination should be diverging, not converging. The sign of equivalent power decides whether the final combination converges or diverges.

Explanation: \( \textbf{Given values:} \) Required equivalent power is \(P_{\text{eq}}=+1\,\text{D}\). One lens has power \(P_1=+4\,\text{D}\). For lenses in contact, \[ P_{\text{eq}}=P_1+P_2 \] Substitute: \[ +1=+4+P_2 \] Rearranging, \[ P_2=+1-4 \] \[ P_2=-3\,\text{D} \] A negative power means the second lens must be diverging. \( \textbf{Final answer:} \) \(-3\,\text{D}\).

Explanation: When two thin lenses are separated by a finite distance, the separation affects the equivalent focal length. The relation is \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1f_2}\), with all lengths written in the same unit. If \(d=0\), the formula reduces to the contact-lens relation \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}\). The separation term is not added with a plus sign in the standard relation. This is why separated lenses cannot always be treated as lenses in contact.

Explanation: \( \textbf{Given data:} \) \[ f_1=+20\,cm,\quad f_2=+30\,cm,\quad d=10\,cm \] For separated thin lenses, \[ \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1f_2} \] Substitute: \[ \frac{1}{F}=\frac{1}{20}+\frac{1}{30}-\frac{10}{20\times30} \] \[ \frac{1}{F}=\frac{3}{60}+\frac{2}{60}-\frac{10}{600} \] \[ \frac{1}{F}=\frac{5}{60}-\frac{1}{60} \] \[ \frac{1}{F}=\frac{4}{60}=\frac{1}{15} \] \[ F=+15\,cm \] \( \textbf{Final answer:} \) \(+15\,cm\), larger than the \(+12\,cm\) value that would come from treating the lenses as touching.

Explanation: \( \textbf{Contact case:} \) \[ \frac{1}{F_c}=\frac{1}{20}+\frac{1}{30} \] \[ \frac{1}{F_c}=\frac{3+2}{60}=\frac{5}{60}=\frac{1}{12} \] \[ F_c=12\,cm \] \( \textbf{Separated case:} \) \[ \frac{1}{F_s}=\frac{1}{20}+\frac{1}{30}-\frac{10}{20\times30} \] \[ \frac{1}{F_s}=\frac{5}{60}-\frac{1}{60}=\frac{4}{60}=\frac{1}{15} \] \[ F_s=15\,cm \] The separation term reduces the net power for this pair of positive lenses. \( \textbf{Final answer:} \) \(12\,cm\) in contact and \(15\,cm\) when separated.

Explanation: \( \textbf{Given data:} \) \[ f_1=+20\,cm,\quad f_2=-50\,cm,\quad d=10\,cm \] Use the separated-lens relation: \[ \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1f_2} \] Substitute: \[ \frac{1}{F}=\frac{1}{20}+\frac{1}{-50}-\frac{10}{20\times(-50)} \] \[ \frac{1}{F}=\frac{1}{20}-\frac{1}{50}+\frac{10}{1000} \] \[ \frac{1}{F}=\frac{5}{100}-\frac{2}{100}+\frac{1}{100} \] \[ \frac{1}{F}=\frac{4}{100}=\frac{1}{25} \] \[ F=+25\,cm \] \( \textbf{Final answer:} \) \(+25\,cm\), so this separated combination is converging.

Explanation: Lens \(L_1\) would form its image \(60\,cm\) to the right of \(L_1\). Lens \(L_2\) is placed \(40\,cm\) to the right of \(L_1\). Therefore the image point due to \(L_1\) lies \(20\,cm\) to the right of \(L_2\). Rays reaching \(L_2\) are still moving left to right and are converging toward that point. For \(L_2\), this point is on the outgoing side of incident light, so it is treated as a virtual object. With the usual sign convention, the object distance for \(L_2\) is \(u_2=+20\,cm\).

Explanation: \( \textbf{Given data for lens } L_2\textbf{:} \) \[ f_2=-10\,cm \] The converging incident rays have a virtual object for \(L_2\): \[ u_2=+20\,cm \] Use the thin lens formula: \[ \frac{1}{f_2}=\frac{1}{v_2}-\frac{1}{u_2} \] Substitute: \[ \frac{1}{-10}=\frac{1}{v_2}-\frac{1}{+20} \] \[ -\frac{1}{10}=\frac{1}{v_2}-\frac{1}{20} \] \[ \frac{1}{v_2}=-\frac{1}{10}+\frac{1}{20} \] \[ \frac{1}{v_2}=-\frac{2}{20}+\frac{1}{20}=-\frac{1}{20} \] \[ v_2=-20\,cm \] \( \textbf{Final answer:} \) \(-20\,cm\), so the final image lies to the left of \(L_2\).

Explanation: In a separated-lens system, the image formed by the first lens becomes the object for the second lens. If the first lens would form its image at a point beyond the second lens, then the rays arriving at the second lens are converging toward a point on its far side. That point is not a real object from which rays are diverging before reaching the second lens. It is treated as a virtual object for the second lens. This is why object distance for the second lens can become positive even though the original object was real.

Explanation: Row P is acceptable because powers add directly for thin lenses in contact. Row Q gives the standard relation for two separated thin lenses. Row R is acceptable because putting \(d=0\) removes the separation term and gives the contact formula. Row S is not acceptable because equivalent focal length is not found by subtracting the separation from the sum of focal lengths. The reciprocal relation must be used because optical power, not focal length itself, combines naturally.

Explanation: \( \textbf{Starting relation:} \) \[ \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1f_2} \] For two fixed lenses, \(f_1\) and \(f_2\) are constants. Let \[ Y=\frac{1}{F} \] Then \[ Y=\left(\frac{1}{f_1}+\frac{1}{f_2}\right)-\left(\frac{1}{f_1f_2}\right)d \] For two convex lenses, \(f_1f_2\) is positive. Therefore the coefficient of \(d\) is negative. \( \textbf{Final answer:} \) a straight line with negative slope.

Explanation: The assertion is true because separation changes how the first lens prepares the rays before they reach the second lens. The reason is also true because the formula for separated lenses includes the term \(-\frac{d}{f_1f_2}\). When \(d=0\), this term becomes zero and the contact relation is recovered. When \(d\neq0\), the equivalent power is generally different from the contact value. The physical separation of lenses is therefore not just a mechanical detail; it changes the optical combination.

Explanation: The separated-lens formula contains reciprocal length terms. For dimensional consistency, \(F\), \(f_1\), \(f_2\), and \(d\) must be expressed in one common length unit. If \(f_1\) and \(f_2\) are in \(cm\), then \(d\) should also be in \(cm\) for that calculation. If powers in \(\text{D}\) are used instead, the separation must be in \(m\). Signs of focal lengths should still be kept because convex and concave lenses contribute differently. Mixing \(m\) and \(cm\) inside the same expression changes the numerical value of the separation term.

Explanation: A prism has two plane refracting faces inclined at an angle called the angle of prism \(A\). A rectangular glass slab has parallel refracting faces, so its emergent ray is parallel to the incident ray when the outside medium is the same on both sides. In a prism, the inclined faces make the emergent ray generally deviate from the original direction. This deviation is the central ray-optics effect studied for a prism. The prism works by refraction at its faces, not by silvered reflection.

Explanation: The angle of prism \(A\) is a geometrical angle of the prism itself. It is measured between the two refracting faces. It should not be confused with the angle of deviation \(\delta\), which is the angle between the original direction of the incident ray and the final emergent ray. The angle of incidence is measured between the incident ray and the normal at the first face. Keeping these angle names separate is necessary before applying prism relations.