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Let, the parallel beam is first incident on convex lens. d = diameter of the beam = 5mm Now, the image due to the convex lens should be formed on its focus (point B) So, for the concave lens, u = +10cm (since, the virtual object is on the right of concave lens) f = -10cm So, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{-10}+\frac{1}{10}=0\Rightarrow\text{v}=\infty$ So, the emergent beam becomes parallel after refraction in concave lens. As shown from the triangles XYB and PQB,$\frac{\text{PQ}}{\text{XY}}=\frac{\text{RB}}{\text{ZB}}=\frac{10}{20}=\frac{1}{2}$
According to the figure, $\frac{\text{x}}{3}=\cot\text{r} \ ...(1)$ Again, $\frac{\sin\text{i}}{\sin\text{r}}=\frac{1}{1.33}=\frac{3}{4}$$\Rightarrow\sin\text{r}=\frac{4}{3}\sin\text{i}=\frac{4}{3}\times\frac{3}{5}=\frac{4}{5} \ \Big(\text{because}\sin\text{i}=\frac{\text{BC}}{\text{AC}}=\frac{3}{5}\Big)$
Both the mirrors have equal focal length f.
The object is placed in the focus of the converging mirror. There will be two images.
Now we have to calculate the image of A and B. Let the images be A',B'. So, length of A'B' = size of image. For A, u = -10cm, f = 6cm Since, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\frac{1}{-10}=\frac{1}{6}$ ⇒ v = 15cm = OA' For B, u = -12cm, f = 6cm Again, $\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{12}$ ⇒ v = 12cm = OB'$\therefore$ A'B' = OA' - OB' = 15 - 12 = 3cm.
$\text{R}_1=+20\text{cm}; \ \text{R}_2=+30\text{cm}; \ \mu=1.6$
Let us assume that it has taken time ‘t’ from A to B.$\therefore \ \text{AB}=\frac{1}{2}\text{gt}^2$
$\text{R}=20\text{cm}, \ \text{f}=\frac{\text{R}}{2}=-10\text{cm}$
Given that, u = distance of the object = -x f = focal length $=-\frac{\text{R}}{2}$ and, V = velocity of object $=\frac{\text{dx}}{\text{dt}}$ From mirror equation, $\frac{1}{-\text{x}}+\frac{1}{\text{v}}=-\frac{2}{\text{R}}$$\frac{1}{\text{v}}=-\frac{2}{\text{R}}+\frac{1}{\text{x}}=\frac{\text{R}-2\text{x}}{\text{R}\text{x}}\Rightarrow\text{v}=\frac{\text{Rx}}{\text{R}-2\text{x}}=$ Image distance
As shown in the figure, OQ = 3r, OP = r So, PQ = 2r For refraction at APB We know, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$$\Rightarrow\frac{1.5}{\text{v}}-\Big(\frac{1}{-2\text{r}}\Big)=\frac{0.5}{\text{r}}=\frac{1}{2\text{r}}$ [because u = -2r]
For the given cylindrical vessel, dimetre = 30cm ⇒ r = 15cm and h = 30cm Now, $\frac{\sin\text{i}}{\sin\text{r}}=\frac{3}{4}\Big[\mu_{\text{w}}=1.33=\frac{4}{3}\Big]$$\Rightarrow\sin\text{i}=\frac{3}{4\sqrt{2}} \ \big[\because \text{r}=45^{\circ}\big]$
Recoil velocity of gun $=\text{V}_{\text{g}}=\frac{\text{mV}}{\text{M}}$ At any time ‘t’, position of the bullet w.r.t. mirror $=\text{V}_{\text{t}}=\frac{\text{mV}}{\text{M}}{\text{t}}=\Big(1+\frac{\text{m}}{\text{M}}\Big)\text{Vt}$ For the mirror, $=\text{u}=-\Big(1+\frac{\text{m}}{\text{M}}\Big)\text{Vt}=\text{kVt}$ v = position of the image From lens formula,$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{-\text{f}}+\frac{1}{\text{kVt}}=\frac{1}{\text{kVt}}-\frac{1}{\text{f}}=\frac{\text{f}-\text{kVt}}{\text{kVtf}}$
Two blocks A and B, each of mass m, are placed on the two sides of the stand. At t = 0, the separation between A and the mirrors is 2 R and also the separation between B and the mirrors is 2 R. The block B moves towards the mirror at a speed v. All collisions which take place are elastic. Taking the original position of the mirrors-stand system to be x = 0 and X-axis along AB, find the position of the images of A and B at t =
