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**Simulator**

**Previous Years AIEEE/JEE Mains Questions**

( 1) $\sin ^{-1}\left(\frac{2}{\sqrt{3}}\right)$

(2) $\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(3) $\sin ^{-1}\left(\frac{1}{2}\right)$

( 4) $\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

** [AIEEE – 2009]**

**Sol.**(2)

(1) 10 m/s

(2) 15 m/s

(3) $\frac{1}{10} \mathrm{m} / \mathrm{s}$

(4) $\frac{1}{15} \mathrm{m} / \mathrm{s}$

** [AIEEE- 2011]**

**Sol.**(4)

$V_{i z f r}=-\left[\frac{f}{f-u}\right]^{2} V_{\text {olrq }}=-\left[\frac{20}{20+280}\right]^{2} \times 15$

(1) Does not depend on colour of light

(2) Increase

(3) Decrease

(4) Remain same

**[AIEEE-2011]**

**Sol.**(2)

$(1)\left(1-\frac{1}{\mu_{1}}\right) \mathrm{h}_{2}+\left(1-\frac{1}{\mu_{2}}\right) \mathrm{h}_{1}$

(2) $\left(1+\frac{1}{\mu_{1}}\right) \mathrm{h}_{1}-\left(1+\frac{1}{\mu_{2}}\right) \mathrm{h}_{2}$

(3) $\left(1-\frac{1}{\mu_{1}}\right) \mathrm{h}_{1}+\left(1-\frac{1}{\mu_{2}}\right) \mathrm{h}_{2}$

$(4)\left(1+\frac{1}{\mu_{1}}\right) \mathrm{h}_{2}-\left(1+\frac{1}{\mu_{2}}\right) \mathrm{h}_{1}$

**[AIEEE- 2011]**

**Sol.**(3)

(1) 5.6 m (2) 7.2 m (3) 2.4 m (4) 3.2 m

**[AIEEE- 2012]**

**Sol.**(1)

(1) 15 cm (2) 20 cm (3) 30 cm (4) 10 cm

**[JEE-Mains- 2013]**

**Sol.**(3)

$\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\left(\frac{3 \times 10^{8}}{2 \times 10^{8}}-1\right)\left(\frac{1}{15}-\frac{1}{\infty}\right)$

f = 30 cm

**[JEE-Mains- 2013]**

**Sol.**(3)

Refer NCERT

(1) $f_{2}>f$ and $f_{1}$ becomes negative

(2) $f_{1}$ and $f_{2}$ both become negative

(3) $f_{1}=f_{2}<f$

(4) $f_{1}>f$ and $f_{2}$ become negative

**[JEE-Mains- 2014]**

**Sol.**(4)

**correct**statement.

(1) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.

(2) The entire spectrum of visible light will come out of the water at various angles to the normal

(3) The entire spectrum of visible light will come out of the water at an angle of $90^{\circ}$ to the normal.

(4) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.

**[JEE-Mains- 2014]**

**Sol.**(4)

Frequency of light () > frequency of green light ($v_{G}$) µ is also greater than $\mu_{\mathrm{G}}$ and critical angle of light is less than green light therefore light will got total internal reflaction and not come out to the air. For frequency of light $(v)<v_{\mathrm{G}}$ ; light will not suffer T.I.R. Therefore light come out to the air

$(1) \theta>\cos ^{-1}\left[\mu \sin \left(A+\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

(2) $\theta<\cos ^{-1}\left[\mu \sin \left(A+\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

(3) $\theta>\sin ^{-1}\left[\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

(4) $\theta<\sin ^{-1}\left[\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

**[JEE-Mains- 2015]**

**Sol.**(3)

Apply Snell’s law at face AB

$1 \sin \theta=\mu \sin \left(r_{1}\right)$

$\theta=\sin ^{-1}\left(\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right.$

for all light transmitted through $\mathrm{AC}, \mathrm{e}<90^{\circ}$

$\Rightarrow \theta>\sin ^{-1}\left(\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right.$

(1) 20 times nearer

(2) 10 times taller

(3) 10 times nearer

(4) 20 times taller

**[JEE-Mains- 2016]**

**Sol.**(4)

Angular magnification is 20.

(1) 1.8 (2) 1.5 (3) 1.6 (4) 1.7

** [JEE-Mains- 2016]**

**Sol.**(2)

(1) real and at a distance of 40 cm from the divergent lens

(2) real and at a distance of 6 cm from the conver

(3) real and at a distance of 40 cm from convergent lens

(4) virtual and at a distance of 40 cm from convergent lens.

**[JEE-Mains- 2017]**

**Sol.**(3)

As parallel beam incident on diverging lens if forms virtual image at $\mathbf{V}_{1}$ = –25 cm from the diverging lense which works as a object for the converging lense (f = 20 cm)

So for converging lens u = –40 cm, f = 20 cm

$\therefore \quad$ Final image

$\frac{1}{\mathrm{V}}-\frac{1}{-40}=\frac{1}{20}$

V = 40 cm from converging lenses.

Very helpful