Physics 1 Marks Question

3. 2%

Explanation:

Illuminance is given by:

$\text{E}=\frac{\text{l}_{\text{o}}\cos\theta}{\text{r}^2}$

$\theta=0^0$

$\frac{\triangle\text{r}}{\text{r}}=1\%$

$\text{E}=\frac{\text{l}_{\text{o}}}{\text{r}^2}$

Differentiating,

$\text{dE}=-2\frac{\text{l}_{\text{o}}}{\text{r}^3}\text{dr}$

As approximation differentials are replaced by $\triangle$,

$\triangle\text{E}=-2\frac{\text{l}_\text{o}}{\text{r}^2}\triangle \text{r}$

$$$\Rightarrow\triangle\text{E}=-2\frac{\text{I}_\text{o}}{\text{r}^2}\Big(\frac{\triangle\text{r}}{\text{r}}\Big)$

$\Rightarrow\triangle\text{E}=-2\text{E}\Big(\frac{\triangle\text{r}}{\text{r}}\Big)$

$\Rightarrow\frac{\triangle{\text{E}}}{\text{E}}=-2\Big(\frac{\triangle{\text{r}}}{\text{r}}\Big)$

$\Rightarrow \frac{\triangle\text{E}}{\text{E}}=-2\times1\%=-2\%$

Since, negative sign implies decrease; hence, illuminance decreases by 2%.

3. $\text{I}\propto \frac{1}{\text{r}^3}$

Explanation:

Let the distance between the parallel straight lines be L.Angle with normal ​$=\theta$

We know,

$\text{I}=\frac{\text{I}_\text{o}\cos\theta}{\text{r}^2}$

From the above figure, we get 

$\text{I}=\frac{\text{I}_\text{o}\sin(90^0-\alpha)}{\text{r}^2}$

$\Rightarrow\text{I}=\frac{\text{I}_\text{o}\sin\alpha}{\text{r}^2}$

$\Rightarrow\text{I}=\frac{\text{I}_\text{o}}{\text{r}^\text{2}}\Big(\frac{\text{L}_\text{o}}{\text{r}\text{}}\Big)$

$\Rightarrow\text{I}=\frac{I_\text{o}}{\text{r}^2}\Big(\frac{\text{L}}{\text{r}}\Big)$

L = constant for parallel moving source

So, I_oL  $=\text{K} \big(\text{constant}\big)$

$\Rightarrow\text{I}=\frac{\text{k}}{\text{r}^3}$

$\Rightarrow\text{I}\alpha\frac{1}{\text{r}^3}$

2. 14 lux

Here,

$\text{r}=\sqrt{2}$

$\tan\theta=\frac{\text{BC}}{\text{AB}}=1$

$\text{I}_\text{o}=40 \text{ lux}$

$\theta=\tan^{-1}(1)=45^0$

The illluminance is given by,

$\text{E}=\frac{\text{l}_\text{0}\cos(45^0)}{\text{r}^2}$

$=\frac{40\times{\cos(45^0})}{(\sqrt{2})^2}$

$=14 \text{ lux}$

3. X_B > X_A, X_B > X_C

Explanation:

Wavelength of light B is 555nm. It has the highest luminosity; hence, X_B will be highest.

Again, 450nm is nearer to 555nm than 700nm.

$\therefore$ 555 - 450 = 105

But 700 - 555 = 145

So, X_A's brightness will be greater than that of X_C.

3. 24s

Explanation:

Here,

$\text{t}_1=12\text{s}$

$\theta_1=0^0$

$\theta_2=60^0$

$\text{t}_2=?$

Let the distance be r.Let the incident luminosity be E_o. 

We have,

$\text{E}_{\theta1}=\frac{\text{E}_\text{o}\cos\theta_1}{\text{r}^2}$

$$$\text{t}_1 \alpha\frac{1}{\text{E}\theta_1}$

$\Rightarrow\text{t}_1=\frac{\text{r}^2\text{k}}{\text{E}_\text{o}\cos\theta_1}$

$\Rightarrow12=\frac{\text{r}^2\text{k}}{\text{E}_\text{o}\cos0}$

$\Rightarrow \frac{\text{r}^2\text{k}}{\text{E}_\text{o}}=12$

Simiolarly,

$\text{t}_2=\frac{\text{r}^2\text{k}}{\text{E}_\text{o}\cos\theta_2}=\frac{12}{\cos(60^0)}$

$\Rightarrow\text{t}_2=12\times2=24\text{s}$

3. $\frac{1}{\text{r}}$

Explanation:

Let us consider two coaxial cylindrical surfaces at distances r and r' from the axis.

Let areas dA and dA' subtend the solid angle $\text{d}\omega$ at the central axis.

The height of the area element will be same, i.e. equal to dy.

Let the breath of dA be dx and that of dA' be dx'.

Now from the arcs,

${\text{dx}}=\text{rd}\theta$

$\text{dx} '=\text{r}'\text{d}\theta$

Now,

$\text{dA}'=\text{dxdy}=\text{rd}\theta\text{dy}$

$\text{dA}'=\text{dx}'\text{dy}=\text{r}'\text{d}\theta\text{dy}$

$\frac{\text{dA}}{\text{dA}'}=\frac{\text{r}}{\text{r}'}$

$\Rightarrow\frac{\text{dA}}{\text{r}}=\frac{\text{dA}^,}{\text{r}^,}=\text{d}\omega$

The luminous flux going through the solid angle d​ω will be:

$\text{dF} = \text{I}\text{​d}\omega$

Now,

$\text{dF}=\text{I}\frac{\text{dA}}{\text{r}}$

If the surfaces are inclined at an angle $\alpha,$

$\text{dF}=\text{I}\frac{\text{dA}\cos{\alpha}}{\text{r}}$

Now, illuminance is defined as 

$\text{E}=\frac{{\text{dF}}}{\text{dA}}=\text{I}\frac{\text{dA}\cos\alpha}{\text{r}}$

$\Rightarrow \text{E}\propto\frac{1}{\text{r}}$

2. 12s

Explanation:

Here,

$\text{d}_1=5\text{cm}= 0.05\text{m}$

$\text{d}_2=10\text{cm}=0.1\text{m}$

$\text{t}_1=3\text{s}$

$\text{t}=?$ 

Let the actual incident illuminance be E_o

Let the iluminance at 3cm distance be E_d1

Let the iluminance at 10cm distance be E_d2

$\cos\theta=1$

$\text{ E}_\text{d1}=\frac{\text{E}_\text{o}}{\text{d}_1^2}$

Now,

$\text{t}_1\alpha\frac{1}{\text{E}_\text{d1}}$

$\Rightarrow \text{t}_1=\frac{\text{k5}^2}{\text{E}_\text{o}}$

$\Rightarrow \frac{\text{k}}{\text{E}_{\text{o}}}=\frac{3}{25}$

Similarly,

$\Rightarrow \text{t}_2=\frac{\text{k10}^2}{\text{E}_\text{o}}$

$\Rightarrow\text{t}_2=\frac{3}{25}\times10^2=12\text{ s}$

Let the illuminance of the candle and the lamp at a distance 1m be I_C and

I_Lrespectively. According to question

$\text{E}=\frac{\text{I}_\text{L}}{(8.0)^2}=\frac{\text{I}_\text{c}}{(0.2)^2}$

$\Rightarrow\frac{\text{I}_\text{L}}{\text{I}_\text{c}}=16$

Again after the wrapping the paper around

$\frac{49}{100}\frac{\text{I}_\text{L}}{(0.8+\text{x})^2}=\frac{\text{I}_\text{c}}{(0.2)^2}$

Either,

x = -0.24

or,

x = -1.36 this value is not acceptable because it is more than 0.8

so, x = 0.24m (-ve sign denotes distance is to be decreased).

x = 24cm

$\therefore$ Illuminance due to the 20cd source.

$\text{E}_1=\frac{20}{(0.4)^2}\ \dots(1)$

$\therefore$ Illuminance due to the 80cd source(E₂) is:

$\text{E}_2=\frac{80}{\text{d}^2}\ \dots(2)$

$\therefore\frac{20}{(0.4)^2}=\frac{80}{\text{d}^2}$

$\Rightarrow\text{d}=0.8\text{m}=80\text{cm.}$

$\frac{\text{I}_1}{\text{I}}=\Big(\frac{80}{20}\Big)^2=16$

$\frac{0.49\text{I}_1}{\text{I}_2}=\Big(\frac{\text{x}}{20}\Big)^2$

$\Rightarrow0.49\times16\times400=\text{x}^2$

$\Rightarrow\text{x}=56\text{cm}$