Physics 5 Marks Questions

1. As shown in figure, for 1^st reflection in M₁, u = -30cm, f = -20cm

$\Rightarrow\frac{1}{\text{v}}+\frac{1}{-30}=-\frac{1}{20}\Rightarrow\text{v}=-60\text{cm}.$

So, for 2^nd reflection in M₂

u = 60 - (30 + x) = 30 - x

v = -x; f = 20cm

$\Rightarrow\frac{1}{30-\text{x}}-\frac{1}{\text{x}}=\frac{1}{20}$

$\Rightarrow \frac{\text{x}-30+\text{x}}{\text{x}(30-\text{x})}=\frac{1}{20}$

$\Rightarrow40\text{x}-600=30\text{x}-\text{x}^2$

$\Rightarrow\text{x}^2+10\text{x}-600=0$

$\Rightarrow\text{x}=\frac{10\pm50}{2}=\frac{40}{2}=20\text{cm}$ or $-30\text{cm}$

$\therefore$ Total distance between the two lines is 20 + 30 = 50cm.

Refractive index $=\mu_1,\mu_2,\mu_3, \ ...\mu_{\text{k}}$

$\therefore$ The shift $\Delta\text{t}=\Big(1-\frac{1}{\mu_1}\Big)\text{t}_1+\Big(1-\frac{1}{\mu_2}\Big)\text{t}_2+ \ ... \ \Big(1-\frac{1}{\mu_{\text{k }}}\Big)\text{t}_{\text{k}} \ ...(1)$

If, $\mu\rightarrow$ refractive index of combination of slabs and image is formed at same place,

$\Delta\text{t}=\Big(1-\frac{1}{\mu}\Big)(\text{t}_1+\text{t}_2+...+\text{t}_{\text{k}}) \ ...(2)$

Equation (1) and (2), we get,

$\Big(1-\frac{1}{\mu}\Big)(\text{t}_1+\text{t}_2+...+\text{t}_{\text{k}})$

$=\Big(1-\frac{1}{\mu_1}\Big)\text{t}_1+\Big(1-\frac{1}{\mu_2}\Big)\text{t}_2+ ...+\Big(1-\frac{1}{\mu_{\text{k}}}\Big)\text{t}_{\text{k}}$

$=(\text{t}_1+\text{t}_2+...+\text{t}_{\text{k}})-\Big(\frac{\text{t}_1}{\mu_1}+\frac{\text{t}_2}{\mu_2}+...+\frac{\text{t}_{\text{k}}}{\mu_{\text{k}}}\Big)$

$=-\frac{1}{\mu}\sum\limits_{\text{i}=1}^\text{k}\text{t}_1=-\sum\limits_{\text{i}=1}^\text{k}\Big(\frac{\text{t}_1}{\mu_1}\Big)\Rightarrow\mu=\frac{\sum\limits_{\text{i}=1}^\text{k}\text{t}_1}{\sum\limits_{\text{i}=1}^\text{k}\big(\frac{\text{t}_1}{\mu_1}\big)}.$

Let a = acceleration of the masses A and B (w.r.t. elevator). From the freebody diagrams,

T - mg + ma - 2m = 0 …(1)

Similarly, T - ma = 0 …(2)

From (1) and (2),

2ma - mg - 2m = 0

$\Rightarrow2\text{ma}=\text{m}(\text{g}+2)$

$\Rightarrow\text{a}=\frac{10+2}{2}=\frac{12}{2}=6$

so, distance travelled by B in t = 0.2 sec is,

$\text{s}=\frac{1}{2}\text{at}^2=\frac{1}{2}\times6\times(0.2)^2=0.12\text{m}=12\text{cm}$

So, Distance from mirror, $\text{u}=-(42-12)=-30\text{cm};\text{f}=+12\text{cm}$

From mirror equation, $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}+\Big(-\frac{1}{30}\Big)=\frac{1}{12}$

$\Rightarrow\text{v}=8.57\text{cm}$

Distance between image of block B and mirror = 8.57cm.

$\text{R}_1=\text{R}_2=10\text{cm}, \ \text{t}=5\text{cm}, \ \text{u}=-\infty$

$\frac{\mu_{\text{g}}}{\text{v}}-\frac{\mu_{\text{a}}}{\text{u}}=\frac{\mu_{\text{g}}-\mu_{\text{a}}}{\text{R}_1}$

$\Rightarrow \frac{1.5}{\text{v}}-\Big(-\frac{1}{\infty}\Big)=\frac{1.5-1}{10}$

$\Rightarrow \frac{1.5}{\text{v}}=\frac{0.5}{10}$

$\Rightarrow\text{v}=30\text{cm}$

$\pi\text{r}^2\text{h}=800\pi+\pi\text{r}_1{^2\text{h}_1}$

$\text{R}_1=+20\text{cm}; \ \text{R}_2=+30\text{cm}; \ \mu=1.6$

1. If placed in air:

$\frac{1}{\text{f}}=(\mu_{\text{g}}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)=\Big(\frac{1.6}{1}-1\Big)\Big(\frac{1}{20}-\frac{1}{30}\Big)$

$\Rightarrow\text{f}=\frac{60}{6}=100\text{cm}$

2. If placed in water:

$\frac{1}{\text{f}}=(\mu_{\text{w}}-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$

$\Rightarrow \Big(\frac{1.6}{1.33}-1\Big)\Big(\frac{1}{20}-\frac{1}{30}\Big)$

$\Rightarrow \Big(\frac{1.60}{1.33}-1\Big)\Big[\frac{1}{6}\Big]$

$=\frac{28}{133\times60}\simeq\frac{1}{300}$

$\Rightarrow\text{f}=300\text{cm}$

$\therefore \ \text{AB}=\frac{1}{2}\text{gt}^2$

$\therefore \ \text{BC}= \text{h}-\frac{1}{2}\text{gt}^2$

1. When the beam is incident on the lens from medium $\mu_{3}$

Then $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$ or $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{(-\infty)}=\frac{\mu_2-\mu_1}{\text{R}}$

or $\frac{1}{\text{v}}=\frac{\mu_2-\mu_1}{\mu_2\text{R}}$ or $\text{v}=\frac{\mu_2\text{R}}{\mu_2-\mu_1}$

Again, for 2^nd refraction, $\frac{\mu_3}{\text{v}}-\frac{\mu_2}{\text{u}}=\frac{\mu_3-\mu_2}{\text{R}}$

or, $\frac{\mu_3}{\text{v}}=-\Big[\frac{\mu_3-\mu_2}{\text{R}}-\frac{\mu_2}{\mu_2\text{R}}(\mu_2-\mu_1)\Big]\Rightarrow-\Big[\frac{\mu_3-\mu_2-\mu_2+\mu_1}{\text{R}}\Big]$

or, $\text{v}=-\Big[\frac{\mu_3\text{R}}{\mu_3-2\mu_2+\mu_1}\Big]$

So, the image will be formed at $=\frac{\mu_3\text{R}}{2\mu_2-\mu_1-\mu_3}$

2. Similarly for the beam from $\mu_3$ medium the image is formed at $\frac{\mu_3\text{R}}{2\mu_2-\mu_1-\mu_3}$

1. In time, $\text{t}=\frac{\text{R}}{\text{V}}$ the mass B must have moved $\Big(\text{v}\times\frac{\text{R}}{\text{v}}\Big)=\text{R}$ closer to the mirror stand

So,

For the block B:

$\text{u}=-\text{R}, \ \text{f}=-\frac{\text{R}}{2}$

$\therefore \ \frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=-\frac{2}{\text{R}}+\frac{1}{\text{R}}=-\frac{1}{\text{R}}$

$\Rightarrow\text{v}=-\text{R}$ at the same place

For the block A:

$\text{u}=-2\text{R}, \ \text{f}=-\frac{\text{R}}{2}$

$\therefore \ \frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=-\frac{2}{\text{R}}+\frac{1}{2\text{R}}=\frac{-3}{2\text{R}}$

$\Rightarrow\text{v}=\frac{-2\text{R}}{3}$ image of A at $\frac{2\text{R}}{3}$ from PQ in the x-direction.

So, with respect to the given coordinate system,

$\therefore$ Position of A and B are $\frac{-2\text{R}}{3},$ R respectively from origin.

2. When $\text{t}=\frac{3\text{R}}{\text{v}},$ the block B after colliding with mirror stand must have come to rest (elastic collision) and the mirror have travelled a distance R towards left form its initial position.

So, at this point of time,

For block A:

$\text{u}=-\text{R}, \ \text{f}=-\frac{\text{R}}{2}$

Using lens formula, v = -R (from the mirror),

So, position xA = -2R (from origin of coordinate system)

For block B:

Image is at the same place as it is R distance from mirror.

Hence, position of image is ‘0’.

Distance from PQ (coordinate system)

$\therefore$ positions of images of A and B are = -2R, 0 from origin.

3. Similarly, it can be proved that at time $\text{t}=\frac{5\text{R}}{\text{v}},$

The position of the blocks will be -3R and $-\frac{4\text{R}}{3}$ respectively.

1. When $\text{t}<\frac{\text{d}}{\text{V}},$ the object is approaching the mirror

As derived in the previous question,

$\text{V}_{\text{image}}=\frac{\text{Velocity of object}\times\text{R}^2}{\big[2\times\text{distance between them}-\text{R}\big]^2}$

$\Rightarrow\text{V}_{\text{image}}=\frac{\text{V}\text{R}^2}{\big[2\big(\text{d}-\text{Vt}\big)-\text{R}\big]^2}$ [At any time, x = d - Vt]

2. After a time $\text{t}<\frac{\text{d}}{\text{V}},$ there will be a collision between the mirror and the mass.

As the collision is perfectly elastic, the object (mass) will come to rest and the mirror starts to move away with same velocity V.

At any time $\text{t}>\frac{\text{d}}{\text{V}},$ the distance of the mirror from the mass will be

$\text{x}=\text{V}\Big(\text{t}-\frac{\text{d}}{\text{V}}\Big)=\text{Vt}-\text{d}$

Here, $\text{u}=-\big(\text{Vt}-\text{d}\big)=\text{d}-\text{Vt}; \ \text{f}=-\frac{\text{R}}{2}$

So, $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{d}-\text{Vt}}+\frac{1}{\big(-\frac{\text{R}}{2}\big)}=-\Big[\frac{\text{R}+2(\text{d}-\text{Vt})}{\text{R}(\text{d}-\text{Vt})}\Big]$

$\Rightarrow\text{v}=-\Big[\frac{\text{R}(\text{d}-\text{Vt})}{\text{R}-2(\text{d}-\text{Vt})}\Big]= $ Image distance

So, Velocity of the image will be,

$\text{V}_{\text{image}}=\frac{\text{d}}{\text{dt}}$ (Image distance)

$=\frac{\text{d}}{\text{dt}}\Big[\frac{\text{R}(\text{d}-\text{Vt})}{\text{R}+2(\text{d}-\text{Vt})}\Big]$

Let, y = (d - Vt)

$\Rightarrow\frac{\text{dy}}{\text{dt}}=-\text{V}$

So, $\text{V}_{\text{image}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\text{Ry}}{\text{R}+2\text{y}}\Big]=\frac{(\text{R}+2\text{y})\text{R}(-\text{V})-\text{Ry}(+2)(-\text{V})}{(\text{R}+2\text{y})^2}$

$=-\text{Vr}\Big[\frac{\text{R}+2\text{y}-2\text{y}}{(\text{R}+2\text{y})^2}\Big]=\frac{-\text{VR}^2}{(\text{R}+2\text{y})^2}$

Since, the mirror itself moving with velocity V,

Absolute velocity of image $=\text{V}\Big[1-\frac{\text{R}^2}{(\text{R}+2\text{y})^2}\Big]$ (since, V = V_mirror + V_image)

$=\text{V}\Big[1-\frac{\text{R}^2}{[2(\text{Vt}-\text{d})-\text{R}^2}\Big]$

1. One due to direct transmission of light through lens.
2. One due to reflection and then transmission of the rays through lens.

$\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{15}+\frac{1}{-40}$

$\Rightarrow\text{v}=24\text{cm}$ (left of lens)

$\Rightarrow\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}}​​\Rightarrow\frac{1}{\text{v}_1}-\frac{1}{-19}=\frac{1}{12}$

$\Rightarrow\text{v}_1=32.57\text{cm}$

$\Rightarrow\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}}​​\Rightarrow\frac{1}{\text{v}_2}-\frac{1}{21}=\frac{1}{12}$

$\Rightarrow\text{v}_2=28\text{cm}$

$\therefore$ Amplitude of vibration of the image is A $=\frac{\text{A}'\text{B}'}{2}=\frac{\text{v}_1-\text{v}_2}{2}$

$\Rightarrow\text{A}=\frac{32.57}{2}=2.285\text{cm}$

$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$

$\Rightarrow\frac{1}{25}=(15-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{-2\text{R}}\Big)=0.5\Big(\frac{3\text{R}}{2}\Big)$

$\Rightarrow\frac{1}{25}=\frac{3}{4}\frac{1}{\text{R}}\Rightarrow\text{R}=18.75\text{cm}$

1. The beam will diverge after coming out of the two convex lens system because, the image formed by the first lens lies within the focal length of the second lens.
2. For 1^st convex lens, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{10} \ (\text{since, u}=-\infty)$

or, $\text{v}=10\text{cm}$

for 2^nd convex lens, $\frac{1}{\text{v}'}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$

or, $\frac{1}{\text{v}'}=\frac{1}{10}+\frac{1}{-(15-10)}=\frac{-1}{10}$

or, $\text{v}'=-10\text{cm}$

So, the virtual image will be at 5cm from 1^st convex lens.

3. If, F be the focal length of equivalent lens,

Then, $\frac{1}{\text{F}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}-\frac{\text{d}}{\text{f}_1\text{f}_2}\Rightarrow\frac{1}{10}+\frac{1}{10}-\frac{15}{100}=\frac{1}{20}$

$\Rightarrow\text{F}=20\text{cm}.$

1. When u = -9.5cm

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{F}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{10}-\frac{1}{9.8}=\frac{-0.2}{98}$

$\Rightarrow\text{v}=-490\text{cm}$

So, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{-490}{-9.8}=50\text{cm}$

So, the image is erect and virtual.

2. When u = -10.2cm

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{F}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{10}-\frac{1}{-10.2}=\frac{102}{0.2}$

$\Rightarrow\text{v}=510\text{cm}$

So, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{510}{-9.8}$

The image is real and inverted.

1. First lens:

u = -15cm, f = 10cm

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\Big(-\frac{1}{15}\Big)=-\frac{1}{10}$

$\Rightarrow\text{v}=30\text{cm}$

So, the final image is formed 10cm right of second lens.

2. m for 1^st lens:

$\frac{\text{v}}{\text{u}}=\frac{\text{h}_{\text{image}}}{\text{h}_{\text{object}}}\Rightarrow\Big(\frac{30}{-15}\Big)=\frac{\text{h}_{\text{image}}}{5{\text{mm}}}$

$\Rightarrow\text{h}_{\text{image}}=-10\text{mm}$ (inverted)

Second lens:

u = -(40 - 30) = -10cm; f = 5cm

[since, the image of 1^st lens becomes the object for the second lens]

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\Big(-\frac{1}{10}\Big)=\frac{1}{5}$

$\Rightarrow\text{v}=10\text{cm}$

m for 2^nd lens:

$\frac{\text{v}}{\text{u}}=\frac{\text{h}_{\text{image}}}{\text{h}_{\text{object}}}\Rightarrow\Big(\frac{10}{10}\Big)=\frac{\text{h}_{\text{image}}}{{-10}}$

$\Rightarrow\text{h}_{\text{image}}=10\text{mm}$ (erect, real).

3. So, size of final image = 10mm

$\mu_2=\frac{3}{2}, \ \mu_1=1, \ \text{u}=0, \ \text{R}=\infty$

1. Let x = distance of the image of the eye formed above the surface as seen by the fish

So, $\frac{\text{H}}{\text{x}}=\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{1}{\mu}$ or $\text{x}=\mu\text{H}$

So, distance of the direct image $=\frac{\text{H}}{2}+\mu\text{H}=\text{H}\Big(\mu+\frac{1}{2}\Big)$

Similarly, image through mirror $=\frac{\text{H}}{2}+(\text{H}\times\text{x})=\frac{3\text{H}}{2}+\mu\text{H}=\text{H}\Big(\mu+\frac{3}{2}\Big)$

2. Here, $\frac{\frac{\text{H}}{2}}{\text{y}}=\mu,$ So, $\text{y}=\frac{\text{H}}{2\mu}$

Where, y = distance of the image of fish below the surface as seen by eye.

So, Direct image $=\text{H}+\text{y}=\text{H}+\frac{\text{H}}{2\mu}=\text{H}\Big(1+\frac{1}{2\mu}\Big)$

Again another image of fish will be formed $\frac{\text{H}}{2}$ below the mirror.

So, the real depth for that image of fish becomes $\text{H}+\frac{\text{H}}{2}=\frac{3\text{H}}{2}$

So, Apparent depth from the surface of water $=\frac{3\text{H}}{2\mu}$

So, distance of the image from the eye $=\text{H}+\frac{3\text{H}}{2\mu}=\text{H}\Big(1+\frac{3}{2\mu}\Big).$

$\text{R}=20\text{cm}, \ \text{f}=\frac{\text{R}}{2}=-10\text{cm}$

$\therefore \ \frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}+\Big(-\frac{1}{3.8\times10^5}\Big)=\Big(-\frac{1}{7.6}\Big)$