NCERT Solutions · Class 12 Physics Ray Optics and Optical Instruments

Objective \(\displaystyle f_o=2.0\ \mathrm{cm}\), eyepiece \(\displaystyle f_e=6.25\ \mathrm{cm}\), separation \(\displaystyle L=15\ \mathrm{cm}\).(a) Final image at \(\displaystyle D=25\ \mathrm{cm}\) (near point):For the eyepiece, the object (the objective's image) must form so the eyepiece gives a virtual image at \(\displaystyle v_e=-25\ \mathrm{cm}\):\[\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\ \Rightarrow\ \frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}=-\frac{1}{25}-\frac{1}{6.25}=-0.04-0.16=-0.20\]\(\displaystyle u_e=-5\ \mathrm{cm}\) — so the objective's real image must form $\displaystyle 5$ cm in front of the eyepiece, i.e. at \(\displaystyle v_o=L-5=15-5=10\ \mathrm{cm}\) from the objective.For the objective:\[\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}\ \Rightarrow\ \frac{1}{u_o}=\frac{1}{v_o}-\frac{1}{f_o}=\frac{1}{10}-\frac{1}{2}=-\frac{4}{10}=-0.4\]\(\displaystyle u_o=-2.5\ \mathrm{cm}\).The object must be placed $\displaystyle 2.5$ cm from the objective.Magnifying power \(\displaystyle M=m_o\times m_e\), with \(\displaystyle m_o=\dfrac{v_o}{u_o}=\dfrac{10}{-2.5}=-4\) and \(\displaystyle m_e=1+\dfrac{D}{f_e}=1+\dfrac{25}{6.25}=1+4=5\):\[M=(-4)(5)=-20\]Magnifying power \(\displaystyle =20\) (magnitude), image inverted.(b) Final image at infinity:Now the objective's image must fall exactly at the eyepiece's focus: \(\displaystyle u_e=-f_e=-6.25\ \mathrm{cm}\Rightarrow v_o=L-f_e=15-6.25=8.75\ \mathrm{cm}\).For the objective:\[\frac{1}{u_o}=\frac{1}{v_o}-\frac{1}{f_o}=\frac{1}{8.75}-\frac{1}{2}=0.1143-0.5=-0.3857\]\(\displaystyle u_o=-2.59\ \mathrm{cm}\).The object must be placed \(\displaystyle \approx2.59\ \mathrm{cm}\) from the objective.\(\displaystyle m_o=\dfrac{v_o}{u_o}=\dfrac{8.75}{-2.59}=-3.375\); for image at infinity, \(\displaystyle m_e=\dfrac{D}{f_e}=\dfrac{25}{6.25}=4\):\[M=(-3.375)(4)=-13.5\]Magnifying power \(\displaystyle =13.5\) (magnitude), image inverted.

Objective: With \(\displaystyle f_o=8.0\,\mathrm{mm}=0.80\,\mathrm{cm}\) and object distance \(\displaystyle u_o=-0.90\,\mathrm{cm}\) (object is beyond \(\displaystyle f_o\), so a real image forms), the lens formula \(\displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) gives \[\frac{1}{v_o}=\frac{1}{f_o}+\frac{1}{u_o}=\frac{1}{0.80}-\frac{1}{0.90}=1.25-1.1111=0.1389\,\mathrm{cm^{-1}}\] \[v_o=7.2\,\mathrm{cm}\] Objective magnification: \(\displaystyle m_o=\dfrac{v_o}{u_o}=\dfrac{7.2}{-0.90}=-8\) (magnitude $\displaystyle 8$).Eyepiece: it must throw a virtual image at the near point, so \(\displaystyle v_e=-D=-25\,\mathrm{cm}\): \[\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\ \Rightarrow\ \frac{1}{-25}-\frac{1}{u_e}=\frac{1}{2.5}\] \[\frac{1}{u_e}=-\frac{1}{25}-\frac{1}{2.5}=-0.44\ \Rightarrow\ u_e=-2.27\,\mathrm{cm}\] The intermediate image (objective's real image) must therefore sit \(\displaystyle 2.27\,\mathrm{cm}\) in front of the eyepiece.Separation \(\displaystyle L=v_o+|u_e|=7.2+2.27=\mathbf{9.47\,cm}\)Magnifying power \(\displaystyle m_e=1+\dfrac{D}{f_e}=1+\dfrac{25}{2.5}=11\) \[M=m_o\times m_e=8\times11=\mathbf{88}\]

For a telescope in normal adjustment, the objective's image lies at its own focus \(\displaystyle f_o\), which coincides with the eyepiece's focus \(\displaystyle f_e\) so the final image forms at infinity.Magnifying power \[m=\frac{f_o}{f_e}=\frac{144\,\mathrm{cm}}{6.0\,\mathrm{cm}}=\mathbf{24}\]Separation (tube length) \[L=f_o+f_e=144+6.0=\mathbf{150\,cm}\]

(a) In normal adjustment, angular magnification of a refracting telescope is \[m=\frac{f_o}{f_e}=\frac{15\,\mathrm{m}}{1.0\times10^{-2}\,\mathrm{m}}=\mathbf{1500}\](b) The Moon subtends a small angle \(\displaystyle \alpha\) at the objective, essentially equal to its angle at the Earth (objective aperture is negligible compared to \(\displaystyle 3.8\times10^{8}\,\mathrm{m}\)): \[\alpha=\frac{d_{\text{moon}}}{r_{\text{orbit}}}=\frac{3.48\times10^{6}\,\mathrm{m}}{3.8\times10^{8}\,\mathrm{m}}=9.16\times10^{-3}\,\mathrm{rad}\] The objective forms its real image at (near) its focus, so the image size is \[d_{\text{image}}=f_o\,\alpha=15\,\mathrm{m}\times9.16\times10^{-3}=0.1374\,\mathrm{m}\] \[\boxed{d_{\text{image}}\approx\mathbf{13.7\,cm}}\]

Use the mirror formula \(\displaystyle \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\) with the Cartesian sign convention: distances measured from the pole; distances of real objects/images (in front of the mirror) are negative, and the focus of a concave mirror is negative while that of a convex mirror is positive. Write \(\displaystyle f=-f_0\) for a concave mirror and \(\displaystyle f=+f_0\) for a convex mirror \(\displaystyle (f_0>0)\), and \(\displaystyle u=-x\) for a real object \(\displaystyle (x>0)\).(a) Concave mirror, object between \(\displaystyle f\) and \(\displaystyle 2f\): here \(\displaystyle f_0<x<2f_0\). \[\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{f_0}+\frac{1}{x}=\frac{f_0-x}{f_0x}\] Since \(\displaystyle x>f_0\), \(\displaystyle f_0-x<0\), so \(\displaystyle v<0\): the image is real (in front of the mirror). Its magnitude is \[|v|=\frac{f_0x}{x-f_0}\] Compare with \(\displaystyle 2f_0\): \(\displaystyle |v|>2f_0 \iff \dfrac{x}{x-f_0}>2 \iff x<2f_0\), which holds by assumption. Hence the real image lies beyond \(\displaystyle 2f\).(b) Convex mirror, any object position: \(\displaystyle f=+f_0\), \(\displaystyle u=-x\), \(\displaystyle x>0\). \[\frac{1}{v}=\frac{1}{f_0}+\frac{1}{x}>0\ \Rightarrow\ v>0\ \text{for every }x>0\] A positive \(\displaystyle v\) (behind the mirror) means the image is always virtual, independent of \(\displaystyle u\).(c) Convex mirror, size and location of that virtual image: \[v=\frac{f_0x}{f_0+x}\ \Rightarrow\ \frac{1}{v}=\frac{1}{f_0}+\frac{1}{x}>\frac{1}{f_0}\ \Rightarrow\ 0<v<f_0\] so the image always lies between the pole and the focus. Magnification \[m=-\frac{v}{u}=\frac{v}{x}=\frac{f_0}{f_0+x}<1\] so it is always diminished, for every object distance \(\displaystyle x\).(d) Concave mirror, object between pole and focus: \(\displaystyle f=-f_0\), \(\displaystyle u=-x\), \(\displaystyle 0<x<f_0\). \[\frac{1}{v}=-\frac{1}{f_0}+\frac{1}{x}=\frac{f_0-x}{f_0x}>0\ \text{(since }x<f_0\text{)}\] So \(\displaystyle v>0\): the image is virtual (behind the mirror). Magnification \[m=-\frac{v}{u}=\frac{v}{x}=\frac{f_0}{f_0-x}>1\ \text{(since }f_0-x<f_0\text{)}\] so the virtual image is enlarged, as required.

A slab of thickness \(\displaystyle t\) and refractive index \(\displaystyle n\), viewed along the normal, makes objects behind it appear shifted toward the viewer by \[\Delta=t\left(1-\frac{1}{n}\right)\] With \(\displaystyle t=15\,\mathrm{cm}\), \(\displaystyle n=1.5\): \[\Delta=15\left(1-\frac{1}{1.5}\right)=15\times\frac{1}{3}=\mathbf{5\,cm}\] The pin appears raised by $\displaystyle 5$ cm. The formula for \(\displaystyle \Delta\) depends only on \(\displaystyle t\) and \(\displaystyle n\), not on where along the $\displaystyle 50$ cm line of sight the slab sits — so no, the answer does not depend on the location of the slab (as long as it stays parallel to the table, between pin and eye).

(a) For light to stay trapped by total internal reflection at the core–cladding interface, the angle of incidence there must be at least the critical angle \(\displaystyle C\), where \[\sin C=\frac{n_2}{n_1}=\frac{1.44}{1.68}=\frac{6}{7}\ \Rightarrow\ \cos C=\frac{\sqrt{13}}{7}=0.5151\] A ray entering the flat end at angle \(\displaystyle i\) refracts to angle \(\displaystyle r\) (measured from the pipe's axis, i.e. the normal to the end face); it then strikes the side wall at angle \(\displaystyle (90^\circ-r)\) from the wall's normal. TIR requires \[90^\circ-r\ge C \ \Rightarrow\ r\le 90^\circ-C \ \Rightarrow\ \sin r\le\cos C\] Applying Snell's law at the entrance face (air, \(\displaystyle n=1\), to core \(\displaystyle n_1=1.68\)): \[\sin i=n_1\sin r\le n_1\cos C=1.68\times0.5151=0.8653\] \[i\le\arcsin(0.8653)\approx\mathbf{60^\circ}\] So all rays incident on the end face within \(\displaystyle 0^\circ\) to about \(\displaystyle 60^\circ\) of the axis undergo total internal reflection inside the pipe.(b) Without the cladding, the core interfaces directly with air, \(\displaystyle n_2'=1\): \[\sin C'=\frac{1}{1.68}=0.5952\ \Rightarrow\ \cos C'=\sqrt{1-0.5952^2}=0.8035\] \[\sin i_{\max}=n_1\cos C'=1.68\times0.8035=1.35>1\] Since this exceeds unity, the condition \(\displaystyle \sin i\le1.35\) is satisfied by every physically possible angle of incidence. So without the outer covering, all rays entering the flat face (any angle from \(\displaystyle 0^\circ\) to \(\displaystyle 90^\circ\)) undergo total internal reflection — there is no restriction on \(\displaystyle i\).

For a real image of a fixed object formed by a convex lens with a fixed object-to-screen distance \(\displaystyle D\), the lens equation restricts \(\displaystyle D\) and \(\displaystyle f\) by \(\displaystyle D\ge4f\); the minimum separation for which a real image is possible at all is \(\displaystyle D_{\min}=4f\), attained when the object and image are symmetric (each at \(\displaystyle 2f\) from the lens). Hence the largest focal length that can still produce a real image over a fixed \(\displaystyle D\) is \[f_{\max}=\frac{D}{4}=\frac{3\,\mathrm{m}}{4}=0.75\,\mathrm{m}\] Maximum possible focal length \(\displaystyle =\mathbf{75\,cm}\).

This is the displacement (conjugate foci) method. If the lens in one position has object distance \(\displaystyle u\) and image distance \(\displaystyle v\) with \(\displaystyle u+v=D=90\,\mathrm{cm}\), by reversibility the second sharp-image position swaps \(\displaystyle u\) and \(\displaystyle v\); the lens shift between the two positions is \[d=v-u=20\,\mathrm{cm}\] Solving \(\displaystyle u+v=90\), \(\displaystyle v-u=20\) gives \(\displaystyle v=55\,\mathrm{cm}\), \(\displaystyle u=35\,\mathrm{cm}\) (taking \(\displaystyle u=-35\,\mathrm{cm}\) with sign convention). Substituting in the lens formula, or equivalently using the standard result of this method, \[f=\frac{D^2-d^2}{4D}=\frac{90^2-20^2}{4\times90}=\frac{8100-400}{360}=\frac{7700}{360}\] \[\boxed{f\approx\mathbf{21.4\,cm}}\]

For the ray to just suffer total internal reflection at the second face, the angle of refraction there equals the critical angle: \(\displaystyle r_2=C\), where \[\sin C=\frac{1}{n}=\frac{1}{1.524}=0.6562\ \Rightarrow\ C=41.0^\circ\] For a prism of refracting angle \(\displaystyle A=60^\circ\): \(\displaystyle r_1+r_2=A\), so \[r_1=A-C=60^\circ-41.0^\circ=19.0^\circ\] Applying Snell's law at the first face (air to glass), \(\displaystyle \sin i=n\sin r_1\): \[\sin i=1.524\times\sin(19.0^\circ)=1.524\times0.3256=0.4962\] \[i=\arcsin(0.4962)\approx\mathbf{29.8^\circ}\] The ray must be incident at about \(\displaystyle 29.8^\circ\) to the normal on the first face.

Principle: thin-lens formula \(\displaystyle \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\), with the lens held close to the eye so the object distance equals the stated viewing distance.Here \(\displaystyle u=-9\ \mathrm{cm}\) (object $\displaystyle 9$ cm from the lens) and \(\displaystyle f=+9\ \mathrm{cm}\) (converging lens).\[\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{9}+\frac{1}{-9}=0 \;\Rightarrow\; v\to\infty\](a) Because the card is exactly at the focal point, the rays emerging from the lens are parallel and the (virtual) image is thrown to infinity. The linear magnification \(\displaystyle m=v/u\) therefore has no finite value — it grows without bound as \(\displaystyle u\to f\). Consequently no finite area can be quoted for the image of a square in this configuration; the image is angularly, not linearly, well defined.(b) With the image at infinity this is the "normal adjustment" of a simple magnifier, for which \[\text{MP}=\frac{D}{f}=\frac{25\ \mathrm{cm}}{9\ \mathrm{cm}}\approx\mathbf{2.8}\] (taking the least distance of distinct vision \(\displaystyle D=25\ \mathrm{cm}\)).(c) No. The linear magnification in (a) is undefined/unbounded, while the magnifying power in (b) is a finite number (≈$\displaystyle 2.8$). This is the sharpest possible illustration that magnification (ratio of image size to object size) and magnifying power (ratio of the angle the image subtends to the angle the object would subtend at the near point) are different quantities — they need not even be comparable, let alone equal, unless the image happens to form exactly at the near point \(\displaystyle D\).

Principle: magnifying power of a simple microscope is largest when the virtual image is pushed in to the near point, \(\displaystyle v=-D=-25\ \mathrm{cm}\), with \(\displaystyle f=9\ \mathrm{cm}\) (the lens of Exercise $\displaystyle 9.22$).(a) \[\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{-25}-\frac{1}{9}=-\frac{34}{225}\ \mathrm{cm^{-1}}\] \[u=-\frac{225}{34}\ \mathrm{cm}\approx \mathbf{-6.62\ cm}\] So the lens (held close to the eye) must be about $\displaystyle 6.6$ cm from the card.(b) \[m=\frac{v}{u}=\frac{-25}{-225/34}=\frac{34}{9}\approx \mathbf{3.78}\] (equivalently, the standard result \(\displaystyle m=1+D/f=1+25/9=34/9\)).(c) Yes, here they are equal. Since the image is formed exactly at the near point (\(\displaystyle |v|=D\)), the angular magnification \[\text{MP}=m\cdot\frac{D}{|v|}=m\cdot\frac{D}{D}=m=\frac{34}{9}\approx \mathbf{3.78}\] Magnification and magnifying power coincide precisely when — and only when — the image is formed at the least distance of distinct vision, because only then is the image angle being compared at the same reference distance \(\displaystyle D\) used to define the unaided-eye angle.

Principle: areal magnification is the square of the linear magnification; then locate the object with the thin-lens formula (\(\displaystyle f=9\ \mathrm{cm}\), same lens).Required area = \(\displaystyle 6.25\ \mathrm{mm^2}\) from an original \(\displaystyle 1\ \mathrm{mm^2}\) square: \[m^2=\frac{6.25}{1}=6.25 \;\Rightarrow\; m=2.5\]With \(\displaystyle v=mu=2.5u\): \[\frac{1}{2.5u}-\frac{1}{u}=\frac{1}{9}\;\Rightarrow\;\frac{-0.6}{u}=\frac{1}{9}\;\Rightarrow\;u=-5.4\ \mathrm{cm}\] \[v=2.5\times(-5.4)=-13.5\ \mathrm{cm}\]The object (card) must be held \(\displaystyle \mathbf{5.4\ cm}\) from the lens, which then places the virtual image \(\displaystyle 13.5\ \mathrm{cm}\) from the eye.No — with the eye very close to the lens, this image sits at only $\displaystyle 13.5$ cm, which is closer than the near point \(\displaystyle D=25\ \mathrm{cm}\). A normal eye cannot accommodate (bring into sharp focus) an object nearer than its near point, so the squares would appear blurred, not distinct.

(a) The angle at the eye is unchanged, but the magnifying glass lets that angle be produced by an object held much closer than the near point \(\displaystyle D\) — closer than the eye alone could ever focus on. Without the lens the object cannot be brought nearer than \(\displaystyle D=25\ \mathrm{cm}\) and still be seen sharply; with the lens it can be held near the focal point (a few cm away) while the eye still sees a sharp virtual image. The "angular magnification" is the gain in subtended angle relative to viewing the same object unaided at the near point — the lens is what makes it legal to bring the object in that close in the first place.(b) Yes, it changes, though only slightly for a small shift. With the object and lens fixed, the image position is fixed; moving the eye back by a distance \(\displaystyle x\) increases the eye-to-image distance to \(\displaystyle |v|+x\), so the angle subtended by the image, \(\displaystyle h'/(|v|+x)\), decreases, and hence \(\displaystyle \text{MP}=m\,D/(|v|+x)\) decreases as the eye is moved away from the lens.(c) Because \(\displaystyle \text{MP}=D/f\) demands an ever smaller, more strongly curved lens as \(\displaystyle f\) shrinks. Such lenses suffer severe spherical and chromatic aberration, and become mechanically hard to grind and hold close to the eye, so image quality collapses well before the formula's magnification could be realised — in practice a simple magnifier is limited to roughly $\displaystyle 8$–10X.(d) The overall magnifying power of a compound microscope is (magnitude) \(\displaystyle M\approx \dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)\), inversely proportional to both focal lengths. A short \(\displaystyle f_o\) lets the objective form a strongly magnified real image close to the eyepiece, and a short \(\displaystyle f_e\) lets the eyepiece then apply a large further angular magnification to that image; both are needed together to reach useful overall magnification.(e) All the light rays collected by the objective and passed on by the eyepiece cross through a small circular region just outside the eyepiece called the eye-ring (exit pupil) — this is where the image-forming beam is narrowest. Placing the eye elsewhere (e.g. right against the eyepiece lens) misses part of this beam and loses field of view/brightness. The eye should be positioned exactly at the eye-ring, a short distance behind the eyepiece — of the order of the eyepiece's own focal length (typically a few mm to about \(\displaystyle f_e\)) — for the widest, brightest view.

Principle: total magnifying power of a compound microscope \(\displaystyle =\) (linear magnification of objective) \(\displaystyle \times\) (angular magnification of eyepiece, image at near point).Eyepiece (\(\displaystyle f_e=5\ \mathrm{cm}\)) at maximum magnifying power (image at \(\displaystyle D=25\ \mathrm{cm}\)): \[\text{MP}_e=1+\frac{D}{f_e}=1+\frac{25}{5}=6\]Required objective magnification: \[m_o=\frac{\text{MP}_{\text{total}}}{\text{MP}_e}=\frac{30}{6}=5\]Objective (\(\displaystyle f_o=1.25\ \mathrm{cm}\)), real inverted image so \(\displaystyle v_o=-5u_o\): \[\frac{1}{-5u_o}-\frac{1}{u_o}=\frac{1}{1.25}\;\Rightarrow\;\frac{-6/5}{u_o}=0.8\;\Rightarrow\;u_o=-1.5\ \mathrm{cm},\quad v_o=7.5\ \mathrm{cm}\]Eyepiece, virtual image at \(\displaystyle v_e=-25\ \mathrm{cm}\): \[\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}=-\frac{1}{25}-\frac{1}{5}=-\frac{6}{25}\;\Rightarrow\;u_e=-\frac{25}{6}\approx-4.17\ \mathrm{cm}\]Set-up: place the object \(\displaystyle \mathbf{1.5\ cm}\) in front of the objective; the objective forms a real, $\displaystyle 5$\(\displaystyle \times\)-magnified image \(\displaystyle 7.5\ \mathrm{cm}\) behind itself, which must fall \(\displaystyle \approx\mathbf{4.17\ cm}\) in front of the eyepiece. So the two lenses should be separated by \[L=v_o+|u_e|=7.5+\frac{25}{6}=\frac{35}{3}\approx\mathbf{11.7\ cm}\]

Principle: astronomical (refracting) telescope magnifying power, \(\displaystyle f_o=140\ \mathrm{cm}\), \(\displaystyle f_e=5.0\ \mathrm{cm}\).(a) Normal adjustment (final image at infinity): \[\text{MP}=\frac{f_o}{f_e}=\frac{140}{5.0}=\mathbf{28}\](b) Final image at the near point \(\displaystyle D=25\ \mathrm{cm}\): \[\text{MP}=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=28\left(1+\frac{5}{25}\right)=28\times1.2=\mathbf{33.6}\]

(a) In normal adjustment the objective's image lies at its own focus, which must also be the eyepiece's focus (so the final image is at infinity): \[L=f_o+f_e=140+5.0=\mathbf{145\ cm}\](b) Tower height $\displaystyle 100$ m at distance $\displaystyle 3$ km subtends a small angle \(\displaystyle \theta=\dfrac{100}{3000}=\dfrac{1}{30}\ \mathrm{rad}\) at the objective. The real image formed by the objective has height \[h'=f_o\,\theta=140\ \mathrm{cm}\times\frac{1}{30}=\frac{14}{3}\approx\mathbf{4.67\ cm}\] (inverted).(c) With the final image pulled in to \(\displaystyle D=25\ \mathrm{cm}\), the eyepiece applies its maximum linear magnification, \(\displaystyle 1+D/f_e=1+25/5=6\), to that objective image: \[h_{\text{final}}=h'\times 6=\frac{14}{3}\times6=\mathbf{28\ cm}\]

Principle: mirror formula \(\displaystyle \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\) applied twice — first to the concave primary, then to the convex secondary, using the fact that the primary's un-formed image acts as a virtual object for the secondary.Primary (concave), \(\displaystyle R_1=220\ \mathrm{mm}\Rightarrow f_1=R_1/2=110\ \mathrm{mm}\). For an object at infinity the primary would form its image \(\displaystyle 110\ \mathrm{mm}\) behind itself — but the secondary mirror, only \(\displaystyle 20\ \mathrm{mm}\) away, intercepts the still-converging beam first. Relative to the secondary, this un-formed point is a virtual object a further \(\displaystyle 110-20=90\ \mathrm{mm}\) behind it: \(\displaystyle u_2=+90\ \mathrm{mm}\).Secondary (convex), \(\displaystyle R_2=140\ \mathrm{mm}\Rightarrow f_2=+R_2/2=+70\ \mathrm{mm}\) (virtual focus, same sign convention).\[\frac{1}{v_2}=\frac{1}{f_2}-\frac{1}{u_2}=\frac{1}{70}-\frac{1}{90}=\frac{9-7}{630}=\frac{1}{315}\] \[v_2=\mathbf{315\ mm}\]The final image forms $\displaystyle 315$ mm from the secondary mirror — equivalently, since the mirrors are $\displaystyle 20$ mm apart, $\displaystyle 295$ mm behind the pole of the primary mirror (through the small central hole typical of a Cassegrain design).

Principle: when a plane mirror rotates through angle \(\displaystyle \theta\), the reflected ray turns through \(\displaystyle 2\theta\) (the incident ray is fixed).Mirror deflection \(\displaystyle \theta=3.5^\circ\), so the reflected ray sweeps \(\displaystyle 2\theta=7^\circ\). On a screen at \(\displaystyle D=1.5\ \mathrm{m}\): \[d=D\tan(2\theta)=1.5\ \mathrm{m}\times\tan(7^\circ)=1.5\times0.1228\] \[\boxed{d\approx 0.184\ \mathrm{m}=\mathbf{18.4\ cm}}\] displaced along the sense of the mirror's rotation.

Principle: an equiconvex lens sitting on a liquid layer over a plane mirror forms an auto-collimating (lens+liquid+mirror) system; the needle's image coincides with the needle when the needle sits at the focus of this combined mirror system, so the measured distance directly gives the equivalent focal length \(\displaystyle F\) each time. For two thin elements of focal length \(\displaystyle f_1,f_2\) in contact with a mirror, light crosses each lens twice, so \(\displaystyle \dfrac{1}{F}=\dfrac{2}{f_1}+\dfrac{2}{f_2}\) (a plane mirror contributes zero power).Liquid removed (glass lens + plane mirror only), \(\displaystyle F=30.0\ \mathrm{cm}\): \[\frac{1}{30}=\frac{2}{f_{\text{lens}}}\;\Rightarrow\;f_{\text{lens}}=60\ \mathrm{cm}\]Radius of curvature of the equiconvex lens (\(\displaystyle n_{\text{glass}}=1.50\), \(\displaystyle R_1=+R,\,R_2=-R\)): \[\frac{1}{60}=(1.50-1)\left(\frac{1}{R}-\frac{-1}{R}\right)=\frac{0.50\times2}{R}\;\Rightarrow\;R=60\ \mathrm{cm}\]With liquid, \(\displaystyle F=45.0\ \mathrm{cm}\): \[\frac{1}{45}=2\left(\frac{1}{60}+\frac{1}{f_{\text{liquid}}}\right)\;\Rightarrow\;\frac{1}{f_{\text{liquid}}}=\frac{1}{90}-\frac{1}{60}=-\frac{1}{180}\] \[f_{\text{liquid}}=-180\ \mathrm{cm}\]The liquid forms a plano-concave lens: its upper surface matches the lens's lower surface (\(\displaystyle R_1=-60\ \mathrm{cm}\)), and its lower surface is flat against the mirror (\(\displaystyle R_2=\infty\)): \[\frac{1}{-180}=(n_{\text{liq}}-1)\left(\frac{1}{-60}-0\right)\;\Rightarrow\;n_{\text{liq}}-1=\frac{60}{180}=\frac13\] \[\boxed{n_{\text{liq}}=\frac{4}{3}\approx\mathbf{1.33}}\]